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Full text of "Algebra and Trigonometry"

OUP -24 -4-4-69—5,01 

OSMANIA UNIVERSITY LIBRARY 

Gall No. 5 / X/ & S& n Accession No. Q % ' S °~ ' 
Author £^''<vt^^y?Ajro^fc<£ ^y/u^ef/ZToh^/i 

This b<#>k should be returned on or before the date last marked below. 



Algebra 

and 

Trigonometry 



by 
ALVIN K. BETTINGER 

Head of Department and 
Professor of Mathematics 
The Crei pli ton University 

and 
JOHN A. ENGLUND 

Formerly Assistant Professor 

of Mat lie mattes 

The Cr eight on University 



INTERNATIONAL TEXTBOOK COMPANY 

Scranton, Pennsylvania 



International Textbooks in Mathematics 

L R. Wilcox 

Professor of Mathematics 
Illinois Institute of Technology 

Consulting Editor 



Second Printing, ] anuary, 1963 



Copyright © I960, by International Textbook Company. All rights reserved. 
Printed in the United States of America by The H addon Craftsmen } Inc., 
at Scr anion, Pennsylvania. Library of Congress Card Number: 60-9987. 



Preface 



The authors believe that in this book the basic material of college 
algebra and trigonometry has been presented with suflicient rigor 
to provide a firm and coherent groundwork for subsequent courses 
in mathematics. The material is presented in such a way that it 
can be grasped by the student without undue assistance. 

The first chapter consists of a number of introductory topics 
which are intended to serve as a review of elementary algebra. 
Actually, something more than a mere review is available in this 
chapter. Not only are the review topics considered from a more 
mature point of view than is usual, but the treatment is inter- 
woven with concepts that are basic for an understanding of more 
advanced mathematical topics. We begin with the algebra of the 
real-number system. Axioms pertaining to fundamental operations 
are given, and the various rules for the elementary operations of 
algebra are derived and logically connected with the basic assump- 
tions. We are led naturally to an ordering of the real-number 
system and to the foundation for a later chapter on inequalities 
that is easier to understand and more useful than the treatment 
one customarily finds in textbooks. 

The second chapter introduces the student to the function con- 
cept, which serves as a basis for much of the remaining work of 
the book. Certain aspects of the discussion become somewhat 
abstract, but the student is reminded that a proper understanding 
of the true nature of a function is important for virtually all later 
courses in mathematics. 

In line with modern demands, the trigonometric functions are 
initially introduced in the third chapter as functions of real num- 
bers. Following this presentation, the transition to functions of 
angles is relatively simple. 

The rest of the volume contains all the usual topics from college 
algebra and trigonometry. In certain instances a particular devel- 
opment may differ somewhat from that usually found. In such 
cases, the authors believe, the departure is to the advantage of 
the student. 



vi Preface 

We are indebted to our colleague, Professor Morris Dansky, for 
his valuable suggestions while the manuscript was in preparation. 
We wish particularly to express our deep appreciation to Pro- 
fessor L. R. Wilcox for his thorough criticism of the manuscript 
and his invaluable suggestions for improvement of the text. Finally, 
a special word of thanks is due the International Textbook Com- 
pany for its cooperation and patience. 

A. K. Bettinger 
J. A. Englund 
Omaha, Nebraska 
August, 1960 



Contents 



1. Introductory Topics 1 

1-1. The Real-Number System 1 

1-2. Fundamental Assumptions 1 

1-3. Operations With Zero 5 

1-4. Reciprocals 5 

1-5. The Real-Number Scale 6 

1-6. Rules of Signs 7 

1-7. Fundamental Operations on Fractions 9 

1-8. Order Relations for Real Numbers 12 

1-9. Absolute Value 14 

1-10. Inequalities Involving Absolute Values 14 

1-11. Positive Integral Exponents 16 

1-12. Algebraic Expressions 17 

1-13. Equations and Identities 18 

1-14. Symbols of Grouping 18 

1-15. Order of Fundamental Operations 20 

1-16. Addition and Subtraction of Algebraic Expressions 21 

1-17. Multiplication of Algebraic Expressions 22 

1-18. Special Products 22 

1-19. Division of Algebraic Expressions 23 

1-20. Factoring 25 

1-21. Important Type Forms for Factoring 27 

1-22. Greatest Common Divisor 30 

1-23. Least Common Multiple 32 

1-24. Reduction of Fractions 33 

1-25. Signs Associated With Fractions 34 

1-26. Addition and Subtraction of Fractions 36 

1-27. Multiplication and Division of Fractions 39 

1-28. Complex Fractions 40 

1-29. Linear Equations 42 

1-30. Linear Equations in One Unknown 43 

2. The Function Concept 49 

2-1. Rectangular Coordinate Systems in a Plane 49 

2-2. Distance Between Two Points 50 

2-3. Functions 52 

2-4. Functional Notation 55 

2-5. Some Special Functions 57 

2-6. Variation 57 

2-7. Classification of Functions 61 



viii Contents 

3. The Trigonometric Functions G3 

3-1. The Point Function P(t) 63 

3-2. Definitions of the Trigonometric Functions (54 

3-3. Identities 68 

3-4. Tables of Trigonometric Functions 71 

3-5. Positive and Negative Angles and Standard Position 75 

3-6. Measurement of Angles 76 

3-7. The Relation Between Radians and Degrees 77 

3-8. Arc Length and Area of a Sector 78 

3-9. Trigonometric Functions of Angles 81 

3-10. Tables of Natural Trigonometric Functions of Angles 82 

4. The Laws of Exponents 86 

4-1. Positive Integral Exponents 86 

4-2. Meaning of a 88 

4-3. Negative Exponents 90 

4-4. Scientific Notation 1)2 

4-5. Rational Exponents 1)2 

4-6. The Factorial Symbol 1)7 

4-7. The Binomial Theorem 1)7 

4-8. General Term in the Binomial Expansion 99 

5. Logarithms 101 

5-1. Definition of a Logarithm 101 

5-2. Laws of Logarithms 102 

5-3. Systems of Logarithms 105 

5-4. Common Logarithms 105 

5-5. Rules for Characteristic and Mantissa 106 



5-6. How to Write Logarithms 108 

5-7. How to Use a Table of Mantissas 108 

5-8. Logarithmic Computation 110 

5-9. Change of Base 113 

6. Right Triangles and Vectors 115 

6-1. Rounding Off Numbers 115 

6-2. Trigonometric Functions of Acute Angles 116 

6-3. Procedures for Solving Right Triangles 117 

6-4. Angles of Elevation and Depression 120 

6-5. Bearing in Navigation and Surveying 121 

6-6. Projections 122 

6-7. Scalar and Vector Quantities 125 

6-8. Logarithms of Trigonometric Functions 131 

6-9. Logarithmic Solution of Right Triangles 133 

7. Trigonometric Functions of Sums and Differences 135 

7-1. Derivation of the Addition Formulas 135 

7-2. The Double-Angle Formulas 139 

7-3. The Half-Angle Formulas 140 

7-4. Products of Two Functions Expressed as Sums, and Sums 

Expressed as Products 143 



Contents ix 

8. Graphs of Trigonometric Functions; Inverse Functions 

and Their Graphs 146 

8-1. Variation of the Trigonometric Functions 140 

8-2. The Graph of the Sine Function 117 

8-3. The Graphs of the Cosine and Tangent Functions 148 

8-4. Periodicity, Amplitude, and Phase 14i) 

8-5. Inverse Functions 155 

8-6. Inverses of the Trigonometric Functions 156 

9. Linear Equations and Graphs 163 

9-1. Solutions of Simultaneous Equations 163 

9-2. Algebraic Solution of Linear Equations in Two Unknowns... 166 

9-3. Linear Equations in Three Unknown* 1(57 

9-4. Graphs of Linear Functions 169 

9-5. Intercepts 170 

9-6. Graphical Solution of Linear Equations in Two Unknowns . . . 171 

10. I JETERMINANTS 173 

10-1 . I )eterminants of the Second Order 173 

10-2. Determinants of the Third Order 175 

10-3. Properties of Determinants 177 

10-4. Solution of Three Simultaneous Linear Equations in Three 

Unknowns 181 

10-5. Systems of Three Linear Equations in Three Unknowns When 

D = () 183 

10-6. Homogeneous Equations 184 

10-7. Sum and Product of Determinants 185 

1 1 . Complex Numbers 189 

The Complex Number System 189 



11-1 

11-2, 

11- 

11-4. 

11-5 

11-6 

11-7. 

11-8, 



The Standard Notation for Complex Numbers 191 

Operations on Complex Numbers in Standard Form 192 

Graphical Representation 195 

Trigonometric Representation 196 

Multiplication and Division in Trigonometric Form 198 

Dc Moivre's Theorem 199 

Roots of Complex Numbers 200 



12. Equations in Quadratic Form 204 

12-1. Quadratic Equations in One Unknown 204 

12-2. Solution of Quadratic Equations by Factoring 20 4 

12-3. Completing the Square 206 

12-4. Solution of Quadratic Equations by the Quadratic Formula . . . 209 

12-5. Equations Involving Radicals 212 

12-6. Equations in Quadratic Form 21 1 

12-7. The Discriminant 215 

12-8. Sum and Product of the Roots 217 

12-9. Graphs of Quadratic Functions 218 

12-J 0. Quadratic Equations in Two Unknowns 221 



x Contents 

12-11. Graphical Solutions of Systems of Equations Involving 

Quadratics 224 

12-12. Algebraic Solutions of Systems Involving Quadratics 227 

12-13. Exponential and Logarithmic Equations 231 

12-14. Graphs of Logarithmic and Exponential Functions 233 

13. Theory of Equations 235 

1 3-1. Introductory Remarks 235 

13-2, Synthetic Division 235 

13-3. The Remainder Theorem 240 

13-4. The Fundamental Theorem of Algebra 241 

13-5. Pairs of Complex Roots of an Equation 243 

13-6. The Graph of a Polynomial for Large Values of a- 244 

13-7. Roots Between a and b If f(a) and f(b) Have Opposite Signs 245 

13-8. Rational Roots 245 

14. Inequalities 248 

14-1. Introduction 248 

14-2. Properties of Inequalities 248 

14-3. Solution of Conditional Inequalities 249 

14-4. Absolute Inequalities 254 

15. Progressions 256 

15-1 . Sequences and Series 256 

15-2. Arithmetic Progressions 260 

15-3. The General Term of an Arithmetic Progression 260 

15-4. Sum of the First n Terms of an Arithmetic Progression 261 

15-5. Arithmetic Means 262 

15-6. Harmonic Progressions 264 

1 5-7. Geometric Progression 265 

15-8. The General Term of a Geometric Progression 265 

15-9. Sum of the First n Terms of a Geometric Progression 266 

15-10. Geometric Means 267 

15-11. Infinite Geometric Progression 268 

15-12. Repeating Decimals 269 

15-13. The Binomial Series 271 

16. Mathematical Induction 273 

16-1. Method of Mathematical Induction 273 

16-2. Proof of the Binomial Theorem for Positive Integral Exponents 275 

17. Permutations, Combinations, and Probability 278 

17-1. Fundamental Principle 278 

17-2. Permutations 279 

17-3. Permutations of n Things Not All Different 280 

17-4. Combinations 281 

17-5. Binomial Coefficients 282 

17-6. Mathematical Probability 283 

17-7. Most Probable Number and Mathematical Expectation 284 

17-8. Statistical, or Empirical, Probability 284 



Contents xi 

17-9. Mutually Exclusive Events 285 

17-10. Dependent and Independent Events 286 

17-11. Repeated Trials 287 

18. Solution of the General Triangle 289 

18-1. Classes of Problems 289 

18-2. The Law of Sines 289 

18-3. Solution of Case I by the Law of Sines: Given One Side and 

Two Angles 290 

18-4. Solution of Case II by the Law of Sines: Given Two Sides and 

the Angle Opposite One of Them 291 

18-5. The Law of Cosines 296 

18-6. Solution of Case III and Case; IV by the Law of Cosines 297 

18-7. The Law of Tangents 298 

18-8. The Half-Angle Formulas 300 

18-9. Area of a Triangle 302 

Appendix 

A. Tables 307 

B. Answers to Odd-Numbered Problems 337 

Index 353 



1 



Introductory Topics 



1-1. THE REAL-NUMBER SYSTEM 

The real-number system that we use in the early part of this 
course is a development from the original counting numbers, or 
positive integers, such as 1, 2, and 3. Almost simultaneously with 
the invention of positive integers, practical problems of measure- 
ment gave rise to positive fractions, such as 1/2, 5/6, and 16/7. 
Much later, in comparatively modern times, the concepts of negative 
numbers and of other types of numbers were gradually developed. 
Negative numbers were invented when the problem of subtracting 
one number from a smaller one presented itself. Thus, the number 
system was soon enlarged to include the negative integers and 
fractions. These positive and negative numbers, together with zero, 
are called the rational numbers. Hence, a rational number is 
defined to be any number that can be expressed as the quotient, or 
ratio, of two integers. For example, —2/3, 5 (which may be con- 
sidered as 5/1), and —7 are rational numbers. 

The number system was then extended to include also numbers 
which cannot be expressed as the quotient of two integers, namely, 
the irrational n}nnbers; examples are y/2 and tt. The two classes of 
numbers, rational and irrational, comprise the real members. These 
numbers are so called in contrast to the imaginary or complex 
numbers considered in Chapter 11. 

1-2. FUNDAMENTAL ASSUMPTIONS 

We shall proceed to introduce the four fundamental operations 
of addition, subtraction, multiplication, and division into the system 
of real numbers. The reader has probably been performing these 
operations in arithmetic and algebra without being conscious that 
certain basic laws were being obeyed. We shall introduce the four 
fundamental operations and state, without proof, the laws or 
assumptions governing them. 

l 



2 Introductory Topics Sec. 1-2 

Addition. It is assumed that there is a mode of combining any 
two real numbers a and 6 so as to produce a definite real number 
called their sum. This mode of combination is called addition. The 
sum of a and b is denoted by a + b. In this sum a and b are called 
terms. 

Multiplication. It is assumed that there is a mode of combining 
any two real numbers a and b to produce a definite real number 
called their product. This mode of combination is called multiplica- 
tion. The product of a and b is denoted by a • b or by ab. The individ- 
ual numbers a and b are called factors of the product. 

Commutative Law for Addition. If a and b are any real numbers, 
then 

(1-1) a + b=b + a. 

Thus 1 , the sum of two numbers is the same regardless of the order 
in which they are added. For example, 

2 + 3=3 + 2. 

Associative Law for Addition. If a, b, c are any real numbers, 
then 
(1-2) (a + b) + c = a + (b + c). 

That is, we obtain the same result whether we add the sum of a and 
6 to c, or we add a to the sum of b and c. Since the way in which 
we associate or group these numbers is immaterial, we may write 
this common value as a + 6 4- c without fear of ambiguity. For 
example, 

2 + 3 + 4 = (2 + 3) + 4 = 2 + (3 + 4). 

Commutative Law for Multiplication. If a and b are any real num- 
bers, then 

(1-3) ab = ba. 

That is, the product of two numbers is the same regardless of the 
order in which they are multiplied. For example, 

2-3 = 3-2. 

Associative Law for Multiplication. , If a, b, c are any real num- 
bers, then 

(1-4) (ab)c = a(bc). 

1 Illustrations of the laws are given here only for the most familiar num- 
bers, the positive integers. It is understood, however, that the laws apply to 
all real numbers. 



Sec. 1—2 Introductory Topics 3 

That is, we obtain the same result whether we multiply the product 
of a and b by c, or we multiply a by the product of b and c. Since 
the way in which we associate or group these numbers is imma- 
terial, we may write the result as abc without fear of ambiguity. 
Thus 

2 • 3 • 4 = (2 • :}) ■ l = 2 • (3 • 4). 

Distributive Law. If a, b, c are any real numbers, then 2 

(1 , r >) a(b + c) = ub + ac. 

This law, which is usually known as the distributive law for multi- 
plication with respect to addition, effects a connection between 
addition and multiplication. The distributive law forms the basis 
for the factoring process in algebra, as will be seen. 
A simple example of the distributive law is 
2- (3 + 4) - 2-3 + 2. 1. 

This law can be extended to the case where the sum consists of 
three or more terms, as in the following illustration : 

:\a(x + 2// - 3;) = liax + 0u// - Uaz. 

For positive integers, multiplication may also be interpreted as 
repeated addition. Thus, by the distributive law, 

3 • 4 = (1 + 1 + 1) • 1 -- (1 • 4) + (1 • 4) + (1 • 4) = 4 + 4 + 4, 

3-4 = 3(1 + 1 + 1 + 1) = (3- 1) + (3- 1) + (3- I) + (3- 1) 
= <*> + - + - + 3. 

Zero. It is assumed that there is a special number called zero 
and denoted by 0, such that, for every real number a, 

(1 (i) a + = a. 

For example, 

3 + 0=3, 0+1=1, + = 0. 

It can be easily shown that only one number with the property 
of can exist. For let 0' be another such number. Then, since 
a + = a and b + 0' = b for any numbers a, b, it follows, by taking 
a -— 0' and b ~ 0, that 

()' + = 0', and + 0' = 0. 

From the commutative law, — 0'. 

2 The right side of (1-5) should read (ah) -f (ac). However, by conven- 
tion, we agree to omit the parentheses when all multiplications are to be 
performed before any addition. 



4 Infroductory Topics Sec. 1—2 

Negative of a Number. It is assumed that for every real number 
a there exists a corresponding number, called the negative of a and 
designated by —a, such that 
(1 7) a + (- „) = 0. 

For example, 

1 + (- 1) =0, (- 2) + 2 =0. 

That each number has but one negative may be shown in the 
following way: Let x be another negative of a, so that a + x — 0. 
Then 

-a = (-<i)+0 = (-„) + (<* + s). 
By associativity, 

- a = ((- a) + a) + x, 
or 

— rt = + x — x. 
In particular, the negative of zero is 

- = - + = 0. 

The Unit. It is assumed that there is a special number called the 
unit and denoted by 1, such that, for every real number a, 

(1 8) a • 1 = a. 

There cannot be a second unit 1'. If there were, we could say that 

1 ■ 1' = 1, 1'. 1 = r, 
whence 1 = 1'. 

Reciprocal of a Number. It is assumed that for every number a 

which is not 0, there is an associated number - , called the rccip- 

a 
vocal of a, such that 

(1-9) «•- = 1. 

a 

The reader may verify the fact that there is only one reciprocal 
of each number. Thus, if x is another reciprocal of a, that is, if 

a • x — 1, then x = - • 
a 

It is important to note the restriction a ^ in the definition of 
the reciprocal. In the next section we shall see why this restriction 
is needed. 

Subtraction. The difference a — b, of any real numbers a and 
b, is defined by 

(1 10) a - b = a + (- 6). 



Sec. 1—4 Introductory Topics 5 

The operation indicated by the si#n minus which produces for any 
two real numbers a and b the real number a — b is called 
subtraction. 

Division. The quotient a f b or 7 or a -r- fr of any real numbers a 

b 
and fr, where b ^ 0, is defined bv 

a-.., ; = «.©• 

The operation associating with real numbers a and fr (fr -V 0) their 
quotient is called division. 

It should be noted that subtraction and division are subordinate 
to addition and multiplication, in that they are defined in terms of 
these latter. The difference a — b is that number .r for which 
b + .v - a. Also, the quotient u b is that number \j for which 
b • y = «. It should be noted that a — a — a + (-«) = for every 
number a, and that r/ 7/ - a • (1, a) — 1 for every number a y 0. 

1-3. OPERATIONS WITH ZERO 

It has already been noted that the special number has the prop- 
erty a + - a for every real number a. In particular, we may let 
a = to obtain 

() + () = 0. 

It has already been noted that —0 = 0, so that a — a 4- = a — 
for every real number a. 

Next, we prove that for every real number a, 

(1-12) a •() -r (). 

Let # — ft • 0. Then, by the distributive law, 

.r = « • = a • (0 + 0) = f/ • + a • - x + x. 

If we add — .r, we obtain 
- :r + (- jc) = U + x ) + (- x ) = x + Or + (- x)) = x + = x. 

Since # = a • 0, (1-12) is established. 

From this last result, it follows that, for b i- 0, 

(113) l = °-(i)=<>- 

1-4. RECIPROCALS 

It was noted that every non-zero number has a reciprocal. We 
can now see why cannot have a reciprocal. If has a reciprocal x, 
then • x = 1. Since it has been shown that • x = 0, we would have 



6 Introductory Topics Sec. 1-4 

to conclude that = 1. However, if = 1 is allowed, then for every 
number a we have 

a = 1 • a — • a — 0. 

Hence, would be the only number in the number system. This 
situation obviously should be ruled out. Therefore, cannot have a 
reciprocal. Moreover, since a/b = a • (1/6), the quotient a/b is not 
defined when 6 = 0. 

The reciprocal of the product of two non-zero numbers can be 
expressed in another way : 

1 1 1 



(1-14) 



a • b ab 



provided that neither a nor b is 0. To prove this result, we begin 
with 

11 7 1 l K , 1 1 

- • i •a«6=--a»T*o= l • 1 =1. 

a b a b 

We then multiply by — r to obtain 

1 1 111 1111 

a • b a * b a b a • b a b a b 

In the preceding proof, free use has been made of the commutative 
and associative laws. 

Finally, if a ^ 0, the reciprocal of the reciprocal of a is a itself. 
Thus, 

(1-15) ^ = a. 

Since 

I7ri -a/«) = i. 

multiplication by f/ gives 

a ~ 1 • a = -7- • (1/a) • a = — r- • 1 = 77 - • 
1/a 1/a 1/a 

1-5. THE REAL-NUMBER SCALE 

Real numbers may be represented by points on a straight line. 
On such a line select an arbitrary point as origin and lay off 
equal unit distances in both directions, as shown in Fig. 1-1. (The 

— 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 

-7 _ 6 -5 -4 -3 -2 -1 I 2 3 4 5 6 7 
Fiu. 1-1 

unit segment may have any length whatsoever.) Label the points 
thus far specified as indicated : is the origin, 1 is the first point to 



Sec. 1—6 Introductory Topics 7 

the right, 2 is the second point to the right, —1 is the first point to 
the left, and so on. Rational numbers that are not integers cor- 
respond to certain other points in a natural way. For example, 1/2 
corresponds to the midpoint of the segment joining points labeled 
and 1; and —7/3 represents the point one-third of the dis- 
tance from the point —2 to the point —3. It is a basic assumption 
concerning the real numbers that every point corresponds to a 
unique real number, and that every real number corresponds to 
exactly one point. The full significance of this assumption cannot 
be developed in an elementary text. 

One observation of importance can be made at this time. The 
non-zero real numbers are divided into two classes. One class con- 
sist of numbers representing points to the "right" of 0, and the 
other consists of numbers representing points to the "left" of 0. 
The first class consists of positive numbers, and the second of 
negative numbers. The number may be considered as constituting 
a third class. It is understood that no two of the three classes — zero, 
positive numbers, and negative numbers — have any numbers in 
common. Thus a number cannot be both positive and zero, both 
positive and negative, or both negative and zero. The specific desig- 
nation of any negative number will include an explicit sign — , which 
is prefixed. (This convention, however, does not exclude the possi- 
bility of allowing a general symbol, such as x, to stand for a nega- 
tive number.) Positive numbers do not require such a sign, 
although frequently the sign f is used. 

It is to be assumed that the sum of two positive numbers is 
positive, as is also the product of two positive numbers. 

1-6. RULES OF SIGNS 

To operate effectively with real numbers, a knowledge of the 
rules of signs and of properties of negative numbers is essential. 
In each of the following relationships, a and b are any two real 
numbers, except that the denominator of a fraction may not be zero. 



(1-16) 




- (- a) = a. 




(1-17) 




- ( a + b) = - a - b. 




(1-18) 




- ( a -b) = - a + b. 




(1-19) 


(- a)b = 


— (ab); in particular, (— \)b = 


- b 


(1-20) 




(-«)(- b) =ab. 




(1-21) 




1 1 
- 6 ~ b' 





8 Introductory Topics Sec. 1-6 

Proofs of (1-16) to (1-28): 
(1-16) - (- a) = - (- a) = a + (- a) + (- (- a)) = a + = a. 

(1-17) - (a + 6) = + - (a + 6) = - a + a + (- b) + b - (a + b) 
= - a-b+ (a + b) - (a + b) 
= -a-b + = -a-b by (1-6), (1-7). 

(1-18) - (a - b) = - (a + (- 6)) = - a - (- b) = - a + 6 

by (1-17), (1-16). 

(1-19) Since (-a)-b + a-6 = (-a + a)-6 = 0-6 = by (1-12), 
(- a) • 6 = (- a) • 6 + a • 6 - (a • 6) = - (a • 6) = - (a • 6). 

(1-20) . (- a) (- 6) = - ((- o) • b) = - (- (c • 6)) 

= ab by (1-19), (1-16). 

d-21) ^ = ^.l=^.6.1=J.. ( - 6) .(.l) 

= l-(-|) = -£ ^ (1-20). 

('■*> :h = -^ = -(-i) = -(-i)--ii 

^£ = (. a) .l = .(a.l) = -J by (1-19). 

(1-23) JZ^ = _ (_ |) = « by (1-22), (1-16). 

, It has already been observed that non-zero numbers are divided 
into two classes, namely, positive and negative. It is assumed that 
if & is positive, then -a is negative ; and that if a is negative, then 
—a is positive. All calculations involving negative numbers can be 
made by performing calculations with positive numbers and apply- 
ing one or more relationships just given. It follows from (1-17), 
for example, that the sum of two negative numbers is negative, and 
is equal to the negative of the sum of the negatives of the given 
numbers. Also, from (1-20) it follows that the product of two 
negative numbers is positive, and is equal to the product of the 
negatives of the given numbers. By (1-19) the product of a posi- 
tive number and a negative number is negative. 



Sec. 1—7 Introductory Topics 9 

1-7. FUNDAMENTAL OPERATIONS ON FRACTIONS 

A further study of the algebra of real numbers leads us to the 
consideration of the fundamental operations as applied to fractions. 
By definition, a fraction is the quotient obtained by dividing 
one number a by another number 6, where b is not zero. We 
call a the numerator and b the denominator; and we generally 
write the fraction a/6, read "a over b" or "a divided by b" 

We shall list the following basic relationships for applying the 
four operations to fractions. In them a, 6, c, d are any real num- 
bers, except that no factor in the denominator of a fraction may 
be zero. 

ac __ a 
bc~b* 



(1-24) 



(1-25) a + b^a + b 



c c 



(1-26) a - + I = 



c d 

a __ b 

c c 
a __b 
c ~d~" cd 



c 


ad + be 


cd 


a — b 


c 


ad — be 



(1-27) 
(1-28) 
(1-29) 

(i-30) ?/£ = l- d - = f 

7 b / d b e be 

A special case of (1-30) is 



a c _ ac 
b'd~W' 



1 



/ c a 
/d = ~c' 



which states that the reciprocal of a fraction is found by inverting 
the fraction. Also, by (1-30), dividing by a fraction is equivalent 
to multiplying by its reciprocal. 



Proofs 


of (1-24) to (ISO): 












(1-29) Since y^ 


= Td b ^- 14 >' 














a c 
b'd '' 


1 1 i 
= a t • c • 3 = (a • 

b d v 


\ 1 1 


= a • 


1 
c 'bd~ 


ac 
bd 


• 


0-24) 




ac _ a c 
be ~ b c 


= ?•! = 


a 
6 






by (1-29). 


(1-25) 


5 + 


b 1 , , 1 

c c c 


= (0 + 6) 


1 
c 


a + b 
c 




.by (1-5). 



10 Introductory Topics Sec. 1-7 

«- 26 > ° + 3 = ^ + ^ = ^ » y(1 -24),(l-25>. 
(1-27) = - + = a • - + (- 6) • - = (a - 6) - 



a — 6 



by (1-22), (1-5). 



(1 _28) 2 » = ? + hfi = i^ + hi^ 

c d c a c • d c • d 



ad — be 



«■*> i/j-H-r;- 



a 1 a ad ad ad 



by (1-22), (1-24), (1-27). 



bl d be T c i c 1 , cd be 

-, b • -= 6 • i • a o 



d d d I'd 

a d 



by (1-29), (1-24). 



Example 1-1. State which of the fundamental assumptions are employed in 
each of the following equations: 

a) 3 +9 + (-5) =3 + (-5) +9. 

b) 11 + (6 + 3) + 7 = (11 + 6) + (3 + 7). 

c) (2 • 3) • 5 = 2 • (3 • 5). 

d) 5(3 + 4) = 5 • 3 + 5 • 4. 

Solution: 

a) The associative and commutative laws for addition. 

b) The associative law for addition. 

c) The associative law for multiplication. 

d) The distributive law. 

Example 1-2. In each of the following, perform the indicated operation: 
a) 2+3. b) (-2) +(-3). c) 5 + (-3). 

d) (+5) -(+3), e) (-7) -(+6). /) (-7) -(-6). 

Solution: Each case can be treated as an addition. 

a) 2 + 3 = 5. 

b) (-2) +(-3) = -5. 

c) 5 + (-3) =2. 

d) (+ 5) ~(+3) = +5 + (-3) = +2. 

e) (-7) -(+6) = -7 + (-6) = -13. 
/) (-7) -(-8) = -7 + <+6) = -1. 

Example 1-3. By using the fundamental assumptions and rules for operations* 
and transforming the left side into the right side, justify the equation 
(a + 6) - (c - d) = (6 - c) + (a + d). 



Sec, 1-7 Introductory Topics 1 1 

Solution: By (1-18), 

(a + 6) - (c - d) = (a + 6) - c + d. 
By (1-1) and (1-2), 

(a + 6) - c + d = (6 - c) + (a + d). 

Example 1-4. By using the fundamental assumptions and rules of operations, 
justify the equation 

6 ^ ac _ be 
a b — c ~ 6 — c 

Solution: By (1-29) and (1-24), 

b ac __ a • (6c) _ 6c 
a 6— c — a* (6— c) ~~ 6 — c 



EXERCISE 1-1 

1. Identify the fundamental law or laws that justify each of the following equations: 
a. x + y — y + x. b. rs = sr. 

c. 2(3 • 5) = (2 • 3)5. d. 5(a + 6) = 5a + 56. 

e. (a -f 2) (6 - 3) = (6 - 3) (a + 2). f. (a + 6)c = c(a + 6) = ca + cb. 

2. Find the value of each of the following: 

a. (-3) +(+5). b. (-5) +(-3). c. (-1) -(-2). 

d. (+7) -(+2). e. (-8) -(-9). f. 0-(-2). 

g. (- 5) - 0. h. 15 + (- 3). i. (- 7) - (- 5). 

j. (-5) -(+5). k. + (-3) -(+4). 1. (+32) -(-23) +(-45). 

3. Evaluate each of the following: 

a. (+2) (-3). b. (-3) (-5). 

c. (-7) (+5). d. (-5) (-9). 

e. 2(-5). f. (-7)0. 

g. (-l)(-2) +(-3)(0). h. (-4) (-5) -(_2)(-l). 

i. (+2) (-3) -(+7) (-5). 

4. Determine the negative of each of the following : 

a. 5. b. - 3. c. 0. d. 2x. 

e. 2/3. f. 2-3. g. 2a - 36. h. - (x - y). 

i. -[(a) (-6)]. j. Sx +2. k. x - 3. I. a + 0. 

5. Find the reciprocal of each of the following: * 
a. ], b. 2/3. c. 3 +|- d. 2 4-i- 

e. 1.02. f. a +6. g. — • h. • 

* 3 a 

i. — ; — ' j. x - 2j. k. * 



x+p J# Jt r-0.1 2-0.3& 



T2 Introductory Topics Sec. 1-7 

6« Prove each of the following equations by using rules for signs and for operations 

with fractions: 

a. a • ( - b) + a • ( - c) = - a • (6 + c). b. - (ac - ad) = a[d -f ( - c)]. 

c. - [6 - (a - c)] = (a - 6) - c. d. 6 /— - h c = a. 

e# . = . f. c [ a ~ ( a - 6/c)] = 6. 

b c - a c — a 

1-8. ORDER RELATIONS FOR REAL NUMBERS 

We shall use the notation a > to express the fact that a is a 
positive number, and the notation a < to indicate that a is nega- 
tive. The symbol > means is greater than, and < means is less 
than. These symbols are called order symbols. 

Assume that a and b are any two given numbers. If a - b > 0, 
we shall write a > b, or b < a, and shall read "a is greater than b" 
or "6 is less than a." As can be easily seen, a > b means that a lies 
to the right of b on the real-number line. When a > b, that is, when 
a - 6 > 0, then 6 — a, which equals -(a - b) by (1-18), is nega- 
tive; and conversely. Hence, a>&(or6<a) if and only if 
b - a < 0. 

The student is familiar with the symbol = (for equality), which 
is used to indicate that two quantities are the same. Thus a = b 
means that the two symbols a and b represent the same mathe- 
matical object. For example, 6 = 3*2. 

If a and 6 are two distinct numbers on the scale, we say "a is 
different from b" or "a does not equal b" and we write symbolically 
a¥=b. The symbol ¥= means does not equal and is called the 
inequality symbol \ 

In general, the oblique line or vertical line through any symbol 
will form a new symbol which is the negation of the original one. 
Thus, a < b means "a is not less than b." In other words, a = 6 or 
a > b (by Property 1 below). For example, 5 < 3. 

Sometimes we shall find it convenient to combine the symbols 
< and = or > and =. We write ^ to mean is less than or equal to, 
and we write ^ to mean is greater than or equal to. 

We thus have order relations on pairs of real numbers, defined 
by either of the following equivalent statements : 

a > b (or b < a) if and only if a — b is positive; 

a > b (or b < a) if and only if b — a is negative. 

The system of real numbers is then said to be ordered by the rela- 
tion > (or the relation <) . Assertions of the type a < b or a > b are 



Sec. 1-8 Introductory Topics 13 

called inequalities. The ordering of the real numbers has the fol- 
lowing properties. 

Property 1. For every pair of real numbers, a and 6, one and 

only one of the following relationships holds : 
a = by or a < b, or a > b. 

Proof of Property 1 : If a = b, the statement is certainly true. 
Saying that a ¥= b is equivalent to saying that a — b ¥= 0, so that 
a - b is either positive or negative. Thus, if a — b is positive, we 
have a > b. If, however, a — 6 is neither positive nor zero, then it 
is negative, and a < b. If two of the three possibilities occurred 
together, we should have, say, a = b and a > b, or a > b and a<b. 
Thus, a — b would be both zero and positive, or both positive and 
negative. Since no overlapping may occur among the three classes 
of numbers, we are thus led to a contradiction. 

Property 2. For any real numbers a, b, c, it is true that 
if a < b and b < c, then a < c. 

Proof of Property 2: If a< b and b < c, then both 6 — a and 
c — b are positive. Let us write c — a as (c - 6) + (b — a). We 
have assumed that the sum of two positive numbers is positive. 
Since c — b and b — a are positive by assumption, their sum, which 
is c — a, is also positive. Hence, c > a, or a < c. 

Property 3. For any real numbers a, b, c, it is true that 

if a > 6, then a + c > b + c. 

Proof of Property 3: By definition, a> b means that a — 6 is 
positive. But, by (1-17), 

(a + c) - (6 + c) = a + c + (-6) + (-c) = a - 6. 

It therefore follows that (a + c) - (b + e) is positive, and that 
a + c > b + c. 

Property 4. For any real numbers a, 6, c, it is true that 
if a > b and c > 0, then ac > 6c. 

Proo/ o/ Property 4: We have assumed that the product of two 
positive numbers is positive. Since both a - b and c are positive, it 
follows that their product is also positive. But (a— b)c = ac — be. 
Therefore, ac — 6c is positive, and ac > be. 

Property 5. For any real numbers a, b, c, it is true that 
if a > b and c < 0, then ac < be. 



14 Introductory Topics Sec. 1-8 

Proof of Property 5: We have noted that the product of two 
numbers with unlike signs is negative. Here a — b is positive, but 
c is negative. Since (a- b)c = ac — bc, and (a - 6) c is negative, it 
follows that ac — bc< 0, or that ac < be. 

According to Property 3, the order symbol in an inequality is not 
changed if the same number is added to or subtracted from both 
sides. It therefore follows that a term on one side of an inequality 
may be transposed to the other side with its sign changed. For 
example, if a — 6 > c, then a > c + b. 

According to Property 5, the order symbol in an inequality is 
reversed if both sides are multiplied or divided by the same nega- 
tive number. 

1-9. ABSOLUTE VALUE 

As a consequence of the properties of the ordering of real num- 
bers, there can be associated with each number a certain non- 
negative number called its absolute value. For any real number a, 
we define the absolute value of a, denoted by \a\, as follows: 

| a | = a, if a ^ 0, and \a\ = — a, if a < 0. 
Thus, | 3 | = 3, since 3 > 0; also J — 3| = — ( — 3) =3, since - 3 < 0. 

1-10. INEQUALITIES INVOLVING ABSOLUTE VALUES 

We shall now consider some inequalities involving absolute 
values. If we let the number x be represented by a point P on a 
number scale, then \x\ is the numerical distance between P and the 
origin. If we let a be a positive number, then \x\ < a means that 
the point P is less than a units from the origin; that is, x lies 
between -a and a. We can write this in the form -a < x and x < a, 
or more briefly in the form — a < x < a. Therefore, the statements 
\x\ < a and —a<x<a mean exactly the same thing. 

A more general inequality which often occurs is \x — b\<a, 
where a > 0. This is equivalent to —a < x — b < a. If 6 is added 
to each term, we may write 6 — a< x < b + a. Hence, the state- 
ments \x — b\<a and 6 — a<x <b + a mean exactly the same 
thing. 

For example, |#-3|<2 may be written -2<#-3<2 and 
means that the distance between x and 3 is less than 2. To solve 
this inequality for x, we add 3 to each term of the inequality, 
obtaining 1 < x < 5. 

The following illustrative examples may help to give a better 
understanding of the processes involved in the solution of the prob- 
lems in Exercise 1-2. 



Sec. 1-10 Introductory Topics 15 

Example 1-5. Arrange the following numbers in increasing order: 
2, - 3.5, 0, 7T, 3.14, |-5|. 

Solution: Since w is approximately 3.1416, the desired order is as follows: 
- 3.5, 0, 2, 3.14, 7T, |-5|. 

Example 1-6. Insert the proper inequality sign (order symbol) between the 
following numbers: 

- 2 and |-2 |. 

Solution: Since | - 2 ] =2, and since - 2 < 2, we have the inequality — 2 

< |-2 |, or |-2 | > -2. 

Example 1-7. Find integers a and b such that a < \/2 < b. 

Solution: Since \/2 may be represented approximately by 1.414, the values 
a = 1 and 6=2 satisfy the inequalities. Thus, 1 < \/2 < 2. Any other pair of 
integers a and b such that a ^ 1 and 6^2 would also satisfy the inequalities. 

Example 1-8. Express the inequality | x j < 3 without using the absolute-value 
symbol. 

Solution: We know that the statements | x | < a and — a < x < a mean exactly 
the same thing. Here a is the positive number 3, and | x \ < 3 means that the 
point represented by x is less than 3 units from the origin; that is, x is between — 3 
and 3. The inequality may be written — 3 < x < 3. 

Example 1-9. Explain the meaning of the inequality | x - 2 | < 1 and write it 
without using the absolute-value symbol. 

Solution: The inequality | x — b \ < a is equivalent to— a < x — b < a. Hence 
| x — 2 | < 1 may be written — 1 < x — 2 < 1. If we add 2 to each term of the 
inequalities, we obtain 1 < x < 3. 

EXERCISE 1-2 

1. Arrange the numbers in each of the following sets in increasing order: 

a. - 3, 0, 4, - 2, 5. b. - 6, - 8, 2, 0, 1/2, - 3/4. 

c. - 2, 10, - 1, - 1/3, - 4. d. - 10, 9, 4, - 3, 3/8, - 6/5. 
e. 3, - 2, 1, V3, - 3/2. f . 1.4. 0, - 2, y/% | - 3 |. 

2. Insert the proper order symbol between the two numbers in each of the 
following : 

*. 3 and 1/3. b. - 3 and | - 3 |. c. y/2 and 1.414. 

d. - 3 and - 2. f e. 22/7 and x. f. 1/8 and 1/6. 

3. Examine each of the following inequalities, and determine whether or not it is 
true. 

a. - 5 > - 3. b. - 3 + 2 < 0. . c | - 3 | > - 3. 

d. t > 22/7. e. | - 2 | < | 2 |. f. | 3 - 7 | > | 5 - 2 |. 



16 Introductory Topics See. 1-10 

4. Find the value of each of the following: 

a.|+2|-|-2|. b.|-3| + |+3|. c|+4|-|-4|. 

d. | - 7 | + | - 5 | - | + 5 |. e. |12 - 4 | - | - 6 |. f. |5-3|+|3|-|2|. 
j. (-18) +3. h.|-9|+|4|. i. -s- 14. 

j. | 4 | -| -5|. 

5. Express each of the following inequalities without using the absolute-value 
symbol: 

a. | x | < 1. b. 5 < 1. c. | x | ^ a. 

d. | 2x | < 4. e. | x - 1 | < 3. f. | x - 1/2 | < 3/2. 

6. In each of the following, find a pair of integers, a and 6, such that the given 
inequalities are satisfied : 

a. a < 5 < b. b. a < - 3 < b. c. a < < b. 

d. a < 7T < 6. e. a < \/3 < 6. f . a < 1 1 - 2 | < 6 

7. If a ^ 3, place the proper order symbol between a + 7 and 10. 

8. If a ^ 5, what can be said about the value of 3a — 2? 



1-11. POSITIVE INTEGRAL EXPONENTS 

If two or more equal quantities are multiplied by one another, 
the product of the equal factors is called a power of the repeated 
factor. Thus 5 2 , read "5 squared," means 5 • 5 ; 5 3 , read "5 cubed," 
means 5 • 5 • 5. In general, a n means the product of n factors each 
equal to a. We call a the base and n the exponent of the power. It 
follows from the associative law that 

a 2 • a 3 = (a • a) (a • a • a) =a,'a'<fa*a = a 5 = a 2+3 . 
Also, if a ** 0, 



a° a • a • a • a* a 



= a • a = a 2 = a 5 "" 3 . 



cr a • a • a 

These and similar results suggest the following laws of expo- 
nents. In them m and n are positive integers. The proofs of these 
laws are reserved for a later chapter. 

Law of Multiplication. To multiply two powers of the same base, 
add the exponents : 

(1-31) a m • a n = a m+n . 

Law of Division. To divide one power of a given base by another 
power of the same base, subtract the exponents : 

Q m 

(1-32) =j = or-*, if o * 0, m > n. 



Sec. 1-12 Introductory Topics 17 

Law for a Power of a Power. To raise a power of a given base to a 
power, multiply the exponents : 

(1-33) (a m ) n = a mn . 

For example, (a 3 ) 2 = a 3 * a 3 = a 3 ' 2 = a 6 . 

Law for a Power of a Product. To obtain a power of a product, 
raise each factor of the product to the given power : 

(1-34) (a&)" = a n b n . 

Thus, (3a 3 ) 2 = 3 2 (a 3 ) 2 = 3 2 a 6 = 9a 6 . 

Law for a Power of a Quotient. To obtain a power of a quotient, 
raise the numerator and the denominator to the given power : 

(1-35) ©"=£' if6 *°- 

fa 2 \ 3 _ (a 2 ) 3 _ a 6 



lhus,if&*0, (^) =^j = — , 



1-12. ALGEBRAIC EXPRESSIONS 

An algebraic expression is formed by combining numbers by 
means of the fundamental operations of algebra. The distinct parts 
of the expression connected by plus and minus signs are called 
terms. The terms of the expression 3x 2 -5xy 2 +7z are 3x 2 y — 5xy 2 , 
and 72. Here the numbers 3, -5, and 7 are called numerical coffi- 
cients, or just coefficients; x 2 , xy 2 , and z are called the literal parts. 

An expression containing one or more terms is called a multi- 
nomial. A multinomial consisting of one term is a monomial. A 
binomial is a multinomial consisting of two terms, and a trinomial 
is a multinomial with three terms. A polynomial is a multinomial 
whose terms are of the form ax m y n z p • • • , where m, n, p, • • • are 
positive integers and a is a numerical coefficient, and where one or 
more of the factors x m , y», z p , • • • may be absent. Thus, 7, 5# 4 , and 

Sxy + 2 are polynomials, while x + - is not. 

y 

The degree of a term of a polynomial is the sum of all the expo- 
nents in its literal part. For example, the degree of Sx 2 is 2, the 
degree of — 5xy 2 is 3, and the degree of lz is 1, because the sums of 
the exponents are, respectively, 2, 3, and 1. 

The degree of a polynomial is the degree of its highest-degree 
term. Thus, in the trinomial 3x 2 -5xy 2 + 7z, the third-degree 
term, ~5xy 2 9 is its highest-degree termf Therefore, Zx 2 - 5xy 2 + 7z 
is a polynomial of the third degree. 



1 8 Introductory Topics Sec. 1-12 

By a polynomial in x of degree n we mean an expression of the 
form 

a x n + a\x n ~ l + • • • + a n , 

where the coefficients a , a u • • • , a n are numerical coefficients, 
a t* 0, and n is a positive integer. If n = 0, we agree that the poly- 
nomial reduces to a number a , which is not 0, and that the degree 
is zero. The number is regarded as a polynomial also, but as one 
having no degree. 

If the typical polynomial just given has degree n, the coefficient 
a is called the leading coefficient. If its leading coefficient is 1, a 
polynomial is called monic. Thus, x* — 2x 2 + bx + 7 is a monic poly- 
nomial of degree 3. 

1-13. EQUATIONS AND IDENTITIES 

An equation is a statement of equality between two numbers or 
algebraic expressions. The two expressions are called members, or 
sides, of the equation. Equations are of two kinds, namely, 
conditional equations and identities. A conditional equation, or 
simply an equation, may be true only for certain values (possibly 
none at all) of the literal quantities appearing. An identity is true 
for all numerical values that can be substituted for the literal 
quantities. 

Illustrations of equations are 

3x - 5 = x + 1 

and 

x 2 - bx + 4 = 0. 

The first one is true only if x = 3, and the second is true only if 
x = 1 or x = 4. 

Illustrations of identities are 

3(x - 2) = 3z - 6 

and 

x 2 — 5x + 4 = (x — 1) (x — 4). 

Each of these equations is true for all values of x. 
1-14. SYMBOLS OF GROUPING 

Parentheses ( ) and other symbols of grouping which have the 
same meaning as parentheses, namely, brackets [ ], braces { }, and 
the vinculum , are used to associate two or more terms which are 
to be combined to form a single quantity. The word "parentheses" 



Sec. 1-14 Introductory Topics 19 

is often used to indicate any or all of these symbols of grouping. 
Removal of the symbols of grouping is accomplished by applying 
the laws of algebra, such as the laws of signs and the distributive 
law. The following examples illustrate the procedure. 

Example 1-10. Remove parentheses from - (2x — 3). 

Solution: The steps may be indicated as follows: 
- (2x -3) = (- 1) (2x -3) 

= (-l)(2s)+(-l)(-3) 
= - 2x + 3. 

Exampl e 1-11. Remove symbols of grouping from Sx — 2[5y + 3 (x — y)] 
— {2y - x - Sy} and collect terms. 

Solution: One way of obtaining the desi red resu lt follows: 
Sx - 2[6y + 3(x - y)] - [2y - x - Zy] 

= Sx - 2[5y + 3x - 3y] - \2y - x + 3y} 
= Sx -2[2y +3x] - {5y - x} 
= Sx - iy - 6x - by + % 
= dx - 9t/. 

The basic rules for enclosing a group of terms in parentheses may 
be stated as follows : 

To write a given expression in parentheses preceded by a plus 
sign, write the terms as they are given, enclose them in parentheses, 
and write + in front of the parentheses. Thus, 

a — b = + (a — b). 

To write a given expression in parentheses preceded by a minus 
sign, change the sign of each term, write the resulting terms in 
parentheses, and write — in front of the parentheses. Thus, 
a — 6 = — ( — a + 6) = — (6-a). 

The first rule is obvious, and the second follows from the rule of 
signs (1-18). 

Example 1-12. 

a) Enclose the last two terms of 2 + Sx - y within parentheses preceded by a 
plus sign. 
6) Enclose these terms within parentheses preceded by a minus sign. 

Solution: 

a) Since the sign before the parentheses is to be +, we enclose 3a - y in its 
given form within the parentheses preceded by a plus sign. In this case the ex- 
pression becomes 2 4- (3x - y). 



20 Introductory Topics Sec. 1-14 

6) Since the sign before the parentheses is to be -, we change the sign of each 
term of Sx — y and enclose — 3# 4- y within the parentheses preceded by a minus 
sign. Then the expression becomes 2 — (- Zx + y). 

1-15. ORDER OF FUNDAMENTAL OPERATIONS 

Parentheses and other symbols of grouping are useful in indicat- 
ing which operation is to be performed first. We have used them 
in this way from the outset. In order to avoid using them unneces- 
sarily, as has been already pointed out, the convention is adopted 
to perform all multiplications first and then the additions (or 
subtractions). If two or more of these symbols of grouping are 
used in the same expression, we usually (though not necessarily) 
remove the innermost pair of symbols first. 

Illustrations in which symbols of grouping are removed follow : 

a) 4 - (6/2) =4-3 = 1. 

b) (4 - 6)/2 = (- 2)/2 = - 1. 

c) 6 + [15/(3 • 5)] = 6 + (15/15) =6 + 1=7. 

d) 6 + (15/3) -5 = 6 + 5-5 = 6 + 25 = 31. 



operations. 



1. 


2«. 


5. 


10». 


9. 


7*. 


13. 


(!)'• 


17. 


!<*'>• 


21. 


a 3 a 4 . 


25. 


(a 2 6 3 ) 2 . 


29. 


(a n b m ) 2 , 


33. 


2l. 





EXERCISE 1-3 




q the group 


from 1 to 36, perform the indicated operation oi 


2. 3*. 


3. (- l) 3 . 


4. 5 3 . 


6. (-2)*. 
10. (-4)'. 


7. (-6) 3 . 

"• S) • 


8. - 10 6 . 


»• (!)'• 


»(-!)• 


16. 4(2°). 


18. |(3«) • 


19. a(5 8 ). 


20. a 2 6(-2) 3 , 


22. (a»)«. 


- 23. (a&) 4 . 


24. a*a 7 . 


26. a -a*- a 3 . 27. a 2 • a* • a 7 . 


28. [(ab)*]*. 


30. (a*& 2 )». 


a 6 
31. V 

a 2 


88.2!. 

a 3 



«■<»■ *©• *©'• 

In each problem in the group from 37 to 48, remove symbols of grouping and 
simplify. 

37. 3 - (6 - 2). 38. x + (y - *). 

39. 4[a + (6 - a)]. 40. (a + 26) - (3a - b). 

41. a - & + (2a - 36). 42. a - [3 + (2a - 4)]. 

43. (a - 26) - (3a + 6). 44. 6s* - [Sx + (2y - x*)]. 



Sec. 1-16 Introductory Topics 21 

45. ax 2 - 2bxy + [(by 2 - 2cx 2 ) - (axy + y 2 )]. 

46. 3n6 - {4ac + [ab - 2ac + ab] - 3a&}. 



47. 2a - [36 + 4c - {3a - b + a - 6 - (3a 4- 2c) } - 4c + a]. 



48. - { - [ - (a - ft - c) - a + (6 - c)]}. 

In each problem from 49 to 60, enclose the last two terms in parentheses. First 
use a plus sign before the parentheses, and then use a minus sign. 
49. a + b + c 50. a 2 - 2aft + b 2 . 51. a 2 - b 2 + c 2 . 

52. x + ?/ - 1. 53. 2a + b - 3c. 54. 3z - 4y + 2z. 

55. x 2 - y 2 - z 2 . 56. - x 2 - ?/ 2 - z 2 . 57. - a 3 ft + aft + ft 2 . 

58. ax 2 - 2a:n/ + y 2 . 59. 2z - 3?/ - 4z. 60. - x 2 - x + 1. 

In each of the following problems, evaluate the given expression. 

61. 16 - (6 - 2). 62. 10 - 6 - 2. 63. (- 3) (-4) - (4) (- 2). 

64. 4 • (6 - 7). 65. 4 • 6 - 7. 66. (4 • 6) - 7. 

67. (3 • 3) - (4 • 2). 68. 3 • 3 - 4 • 2. 69. 3 • (3 - 4) • 2. 

70. 3 • 3 - (4 • 2). 71. (8/2) + 4. 72. 8/(2 + 4). 

73. 8 + (4/2). 74. (8 + 4)/2. 75. 8 + 4(1/2) + 3. 

76. (8 + 4)/ (2 + 3). 77. 8 + (4/2) + 3. 78. 8 + ((4/2) + 3). 

"«• »(S?^/»)->- "•l^/c-"- 

-[f^-OA 



1-16. ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSIONS 

Terms, such as 2x 2 y 3 and 5x 2 y s , which have the same literal parts, 
are called similar or like terms and may be added or subtracted by 
adding or subtracting their coefficients. To illustrate, let us con- 
sider an example. 

Example 1-13. Add 5xy 2 , 7x 2 y, - 2xy 2 , - 9x 2 y, 4x 2 y*. 

Solution: Collecting like terms and adding coefficients, we have 

(5 - 2)xy 2 + (7 - 9)x 2 y + 4x 2 t/ 3 = 3xy 2 - 2x 2 y + 4x*y*. 

The procedure used in the solution of Example 1-13 follows at 
once from the distributive law. For example, the sum of the terms 
5xy 2 and -2xy 2 is obtained as (5 - 2)xy 2 or Sxy 2 . 

This leads at once to the rule for the addition (or subtraction) of 
algebraic expressions. In practice, we usually arrange like terms 
in vertical columns, and then we find the sum of each column by 
prefixing the sum of the numerical coefficients in the column. The 
procedure may be made clearer by means of the following examples. 



22 Introductory Topics Sec. 1-16 

Example 1-14. Add 2x 2 - Sxy + z,x 2 - 5z, 2xy + 3z. 

Solution: 

2x 2 - Zxy -f z 

x 2 - 5« 

2xy + 3z 

3a; 2 — xy — z. 

Example 1-15. Subtract 2a 2 - 36 + c 2 from 3a 2 - c 2 . 

Solution: 3a 2 — c 2 

2a 2 - 36 + c 2 



a 2 + 36 - 2c 2 . 



1-17. MULTIPLICATION OF ALGEBRAIC EXPRESSIONS 

With the help of the distributive law for multiplication, the 
product of two algebraic expressions is found by multiplying each 
term of one by each term of the other and combining like terms. 

Example 1-16. Multiply 3x 2 - 2xy + y 2 by 2x - 3y. 
Solution: 



3x 2 - 2xy + y 2 
2x - Sy 




6x* _ ± X 2 y + 2xy 2 
- 9x 2 y + 6xy 2 - 


-3^ 


6z 3 - lSx 2 y + &rz/ 2 


-32/3. 



1-18. SPECIAL PRODUCTS 

The following typical forms of multiplication occur so frequently 
that we should learn to recognize them quickly and to obtain the 
products without resorting to the general process of multiplication. 
They should be learned thoroughly. 

(1-5) a(b + c) = ab + ac. 

(1-36) (a + b) (a - 6) = a 2 - 6 2 . 

(1-37) (a + b) (a 2 - ab + b 2 ) = a 3 + 6 3 . 

(1-38) (a - 6) (a 2 + ab + b 2 ) = a 3 - 6 3 . 

(1-39) (a + 6) 2 = a 2 + 2a6 + fe 2 . 

(1-40) (a - 6) 2 = a 2 - 2a& + 6 2 . 

(1-41) (ax + by) (ex + dy) = acz 2 + (ad + bc)xy + bdy 2 . 



See. 1-19 Introductory Topics 23 

1-19. DIVISION OF ALGEBRAIC EXPRESSIONS 

To divide a polynomial by a monomial, divide each term of the 
polynomial by the monomial and add the results. 

This rule follows immediately from the rule for fractions 
expressed by (1-25). 

Example 1-17. Divide 6a 2 x 2 - 12a% - 30a«x* by 15a 3 x 2 . 

6a 2 x 2 - 12a 4 x - 30a 6 z 5 6a 2 z 2 12a 4 z 30a 6 z 5 



Solution: 



15a*x 2 15a 3 x 2 15a 3 # 2 15a 3 x 2 

= f _ |a _ 2a3x3 _ 
ba ox 



To divide a polynomial (the dividend) by a polynomial (the 
divisor) , arrange both according to descending or ascending powers 
of some common literal quantity. Then proceed as follows: 

Divide the first term of the dividend by the first term of the 
divisor to obtain the first term of the quotient. 

Multiply the entire divisor by the first term of the quotient, and 
subtract this product from the dividend. 

Use the remainder found by this process as a new dividend, and 
repeat the process. Continue the work until you obtain a remainder 
that is of lower degree in the common literal quantity than the 
divisor. 

Example 1-18. Divide 6a; 3 - 5x 2 + 3x + 1 by 2x - 1. 

Solution: 2s - 1 | 6x 3 - 5x 2 + 3x + 1 1 3x 2 - x + 1 

6s 2 - 3s 2 

- 2x 2 + 3x 
-2x 2 +x 



2x + 1 
2x - 1 

2. 

The division can be checked by finding the product of (2x - 1) and (3x 2 - x -f 1) 
and adding 2, proving that 6x 3 - bx 2 + 3x + 1 = (2x - 1) (3z 2 - x + 1) + 2. 

Any problem in division, in general, may be checked by means df 
the relationship ? 

dividend = (divisor) (quotient) + remainder. 
This equation is an identity ; that is, it is true for all values of the 
literal quantities. Indeed, this equation supplies the underlying 
meaning of the process of division. 



24 Introductory Topics Sec. 1-19 

EXERCISE 1-4 

In each problem in the group from 1 to 12, add the given expressions which are 
separated by commas. 

1. - xy, Sxy. 2. ix 2 y 2 , - 2xy. 3. Sx 2 y 2 , - 2x 2 y 2 . 

4. 6a +116, - 3a + 26. 5. 4a 2 - 2b 2 , a 2 - 46 2 . 6. 3a - 2b -f- 4c, 4a + 36 - 6c. 

7. — 3a + b — c, — a — & 4- c. 

8. 3x -b 2?/ - z, x + ?/ - Sz, 4x - 3y -f 2z. 

9. x 2 - 2xy + ?/ 2 j 4x?/, - t/ 2 . 

10. 3x 2 - 4xy 2 + 2?/ 3 , 4x 2 y - 2x 2 - y 3 , 4xy 2 - 2?/ 3 . 

11. 2x 3 - 3x -f 1, x 2 4 2x - 3, 2x 2 - x 3 + 4 - 2x. 

12. 4ax 4- 3bxy - Ax 2 , 2bx - bx 2 4 2axy, Sx - 2y. 

In each problem from 13 to 24, subtract the second expression from the first. 
13. — xy, Sxy. 14. 4x 2 ?/ 2 , — 2xy. 

15. Sx 2 y 2 , - 2x 2 y 2 . 16. 6a + 116, - 3a 4- 26. 

17. 4a 2 - 26 2 , a 2 - 46 2 . 18. 3a - 26 4 4c, 4a + 36 - 6c. 

19. - 3a + 6 - c, - a - 6 + c. 20. x 3 + x 2 4- x 4 1, x 3 - x 2 - x -1. 

21. x 2 -f 2xy 4 2/ 2 , 4x?/. 

22. 3 - 2x 4 x 2 - x s + 3x 4 - 2x 5 , 3x 5 4 4x 4 - x 3 -f- 2a: 2 - 2x 4 1 . 

2 12 2 2 17 

23. -a; 3 - -a; 2 -4- ^xy - -y 2 , -x 3 4 jxy - ^/ 2 ' 

24. -xy 2 - =x 2 y 4- -xy 2 - ^y 2 > ^x 2 y - -xy 4- ^/ 2 • 

In each problem from 25 to 60, perform the indicated operations. 

25. (6x)(-3?/). 26. (4x)(-2). 

27. (5x 2 y) (- 2xy). 28. (3a) (26 2 ) (- 4c). 

29. (- 3a6) (46c) (- 2a 3 ). 30. (x - y)2a. 

31. (3x + 2y) (- 2x). 32. (6 - 3a) (2). 

33. (4a5|/ 4 2y 2 ) (3xy). 34. (4a 2 + 66 2 ) (Sab). 

35. (2x + Sy) (x - y). 36. (3a - 6) (a + 26). 

37. 4x 2 (x 2 - 2xy - y 2 ). 38. Sxy(2x 2 y + Sxy - 4y 2 ). 

39. (x 2 + xy - y 2 ) (x - y). 40. (x 2 - y 2 ) (x + y) (x - y). 

41. 4x/(- 2). 42. 3x 2 y/2xy. 

43. 4x 2 y 3 z/2xy 2 *. 44. 20x 6 y 4 2yi0x 6 yz 3 . 

45. 4a6/(- 4a6). 46. (2xy) 2 /2xy 2 . 

'47. (2xy) 2 /2(xy) 2 . 48. Or 2 -2xy)/(-x). 

49. (3x?/ 2 - 6xy + 9y 2 )/(- 3y). 50. (a: 2 ?/ 3 - 3x 2 y 2 + 4x 3 y 2 )/x 2 y 2 . 

51. (x 2 + 6x + 5)/(x + 1). 52. (9x 2 - 6x + l)/(3x - 1). 

53. (x 2 - y 2 )/(x - y). > 54. (x - y) 2 /(x - #)• 

55. (x 4 - y 4 )/(x - y). 56. (x 3 4 3x 2 4 Sx + l)/(x + l) 2 . 

57. (x 3 + 2/ 3 )/(x 4- j/). 58. (z + y)*/(x + y). 

59. (x 3 - y 3 )/(x - y). 60. (z 4 + x 3 + 3x 2 + 2x + l)/(x 2 + 2). 

61. Divide x 2 - y 2 - (x - y) 2 - (x - y) (x y) by x - y. 

62. Divide (a + 6) 2 + 6(a 4- b) + 5 by a -f 6 + 5. 

63. Divide x 4 + 4x 3 4- 6x 2 4- 4x + 1 first by x 4- 1 and then by x 2 + 2x 4- 1. 

64. Multiply x 2 - 2x - 3 by x + 4, and divide the result by x 4- 1. 



Sec. 1-20 fnfrocfucfory Topics 25 

65. Divide x 5 + x 4 + 3x 3 - 2# 2 - 3 by x - 1 and add the quotient to the excess of 
3z 4 4- 2x 2 - 9x + 7 over 2z 4 + 2x 3 + 7x 2 + 3z + 4. 

66. a) Under what conditions will ( - #) n be positive? 6) When will it be negative? 
Assume first that x is positive and then that x is negative. 

67. For what values of n will (— a) n be equal to — a n ? 

1-20. FACTORING 

Factoring a quantity is the process of finding quantities which, 
when multiplied together, yield the given quantity. When a quan- 
tity A is expressed as a product B • C, B and C are called factors 
or divisors of A and are said to divide A. Also, A is called a multiple 
of each of Z? and C. These concepts are applicable to numbers or 
algebraic expressions generally, but are most useful when restricted 
to apply to integers or to polynomials. Such restriction will be 
adhered to in this book. Thus, when an integer is to be factored, 
the factors sought are to be integers. And when a polynomial is to 
be factored, the desired factors are to be polynomials. 

Let us first review the fundamentals of factoring integers (posi- 
tive, negative, or zero) . First, it is clear that every integer n may 
be expressed as 1 • n or (-l)-(-w). Such factorizations are called 
trivial. If an integer n, other than 4-1 or — 1, has no factorizations 
other than trivial ones, then n is called a prime (number). An 
integer having a non-trivial factorization is called composite. 
Examples of prime integers are 3, 7, and —11; examples of com- 
posite integers are 6 and —40. 

Let n be a composite integer. Then a non-trivial factorization 
m 'p exists in which \m\ and \p\ are less than \n\ and greater than 
1. If both m and p are primes, then n is expressed as a product of 
primes. If not, at least one of m and p, say p, is composite, and so 
p = r • s. Hence, n~ m • r • s. The process begun may be continued 
if any one of m, r, and s is composite, and additional factors may 
be found, until the process cannot be continued further, in which 
case only prime factors are obtained. We may conclude that the 
factoring process must terminate, since at any stage the new fac- 
tors introduced are numerically less than their product. The 
fundamental theorem of arithmetic guarantees that every com- 
posite integer is a product of primes, which &re unique except for 
their signs or the order in which they are written. For example, 
successive factoring of 156 gives 

156 = 39 • 4 = 13 • 3 • 4 = 13 • 3 • 2 • 2, 
and the final factors 13, 3, 2, 2 are primes. Another valid factor- 
ization of 156 into primes is as follows : 

150 = 2-3- (- 13) -(-2). 



26 Introductory Topics Sec. 1-20 

Let us turn now to polynomials with real coefficients. These may 
be polynomials in x, such as x 2 — 1 ; or polynomials in x and y, 
such as x 2 + Sxy + 2y 2 ; or, in fact, polynomials in any number of 
literal quantities. Every polynomial F has a factorization of the 
form 

F = l(aF), 

for every non-zero number a. And, of course, the factors 1/a and 
aF are themselves polynomials. Such factorizations are called 
trivial. If a polynomial F has no factorizations other than trivial 
ones, then F is called a prime polynomial, or an irreducible poly- 
nomial. A polynomial having a non-trivial factorization is called 
composite or reducible. 

Examples of prime polynomials are Sx + 2, 2x + 2y y and x 2 + 3. 
Every polynomial in x of the first degree may be shown to be 
prime; and certain polynomials of higher (even) degree in x also 
are prime. A development of criteria for primeness of polynomials 
lies beyond the scope of this book. 

Examples of composite polynomials are x 2 — 4, xy + y 2 and 
xz + yz + xu + yu, because each of these has a non-trivial factor- 
ization. Thus, we have 

x 2 - 4 = (x + 2) (x - 2), 
xy + ij 2 = (x + y)y, 
xz + yz + xu + yu = (x + y) (z + u). 

Let / be a composite polynomial in x with real coefficients, not all 
of which are zero. Then it can be shown that there exist poly- 
nomials g and h, the degree of each of which is less than that of /, 
such that 

f = g-k. 

If g and h are primes, then / is a product of primes. If not, we may 
proceed to factor (non-trivially) one or both of g and h, and we 
can continue this process until primes are obtained. The process 
must terminate eventually, since each non-trivial factorization 
leads to polynomials of lower degree. 

The argument just presented applies only to polynomials in one 
literal quantity x. However, the principle may be extended to 
apply to polynomials in any number of literal quantities x, y, z, 
• • • . Thus, 

/ = Pip 2 --'Pn, 

where p lf p 2 , • • • , p n are prime polynomials. The problem of carry- 



Sec. 1-21 Introductory Topics 27 

ing out actual factorizations for certain types of polynomials is 
considered in the next section. 

A special class of polynomials deserves particular attention. 
This is the class that consists of polynomials in which all the 
numerical coefficients are integers. It is possible to prove a factor- 
ization theorem such as that just stated, but yielding factors which 
are polynomials having only integral coefficients. When the general 
theorem can yield prime factors all of which have integral coeffi- 
cients, the two theorems give the same result. Otherwise, they 
will give different results. 

For example, both theorems applied to x 2 - 4 yield the prime 
factorization (x 4-2) (x -2). The polynomial Sx 2 - 4 has no non- 
trivial factors with integral coefficients. However, when other 
real coefficients are allowed, we have 

3x 2 - 4 = (y/3x + 2) (y/Zx - 2), 

in which the coefficient \/S is a perfectly acceptable real number. 
Again, when factors with integral coefficients are desired in such a 
case as 36# + 24ty, we shall agree to remove and factor common 
numerical factors, to obtain 

36x + 24y = 3 • 2 • 2 • (3x + 2y). 

Here the prime factors are the numerical primes 3, 2, 2 and the 
prime polynomial 3x + 2y. 

In what follows, whenever a polynomial has only integral coeffi- 
cients, we agree to restrict ourselves to factors of the same kind. 
In similar fashion, when the given polynomial has only rational 
coefficients, we shall search for factors having only rational 
coefficients. 

1-21. IMPORTANT TYPE FORMS FOR FACTORING 

The equations in Section 1-18 applied "in reverse" are formulas 
for factoring. Success in factoring a polynomial therefore depends 
on ability to recognize the polynomial as being a particular type of 
product and as having factors of a definite form. Verify the follow- 
ing type forms by carrying out the indicated multiplications and 
learn each form. 

Type 1: Common Monomial Factor, From (1-5) we have 

(1-42) ab + ac = a(b + c). 

Example 1-19. Factor 4x 2 y - 6xy 2 . 
Solution: ±x 2 y - Qxy 2 = 2xy(2x - Zy). 



28 Introductory Topics Sec. 1-21 

Instructions for actually removing the monomial factor in 
Example 1-19 may be put this way : Write the common factor, 2xy, 
and in parentheses following 2xy write the algebraic sum of the 
quotients obtained by dividing successively every term of 4x 2 y - 
6xy 2 by 2xy, in accordance with the distributive law. 

Type 2: Difference of Two Squares. From (1-36) we have 

(1-43) a 2 - b 2 = (a + b) (a - b). 

Example 1-20. Factor 9x 2 - 25s/ 2 . 

Solution: 9x 2 - 2hy 2 = (3x + by) (3x - by). 

Type 3: Sum and Difference of Two Cubes. From (1-37) and 
(1-38) we have 

(1-44) a 3 + 6 3 = (a + b) (a 2 - ab + b 2 ) 

and 

(1-45) a 3 - 6 3 = (a - b) (a 2 + ab + b 2 ). 

Example 1-21. Factor &r 3 - 27y*. 

Solution: 8x* - 27 y* = (2x - Zy) [(2a;) 2 + (2s) (Zy) + (3?/) 2 ] 
= (2x - Sy) (4z 2 -f 6x2/ + 9y 2 ). 

Type 4: Perfect-Square Trinomials. From (1-39) and (1-40), 

(1-46) a 2 + 2ab + b 2 = (a + 6) 2 

and 

(1-47) a 2 - 2a6 + b 2 = (a - 6) 2 . 

Example 1-22. Factor 4x 2 + 12xy 2 + 9y 4 . 

Solution: Ax 2 + 12;n/ 2 -f 9?/ 4 = (2a;) 2 4- 2(2s) (3z/ 2 ) + (3y 2 ) 2 

= (2* + Sy 2 ) 2 . 

Type 5: General Trinomial. Factorization of the general tri- 
nomial may be indicated as follows : 

(1-48) Ax 2 + Bxy + Cy 2 = (ax + by) (ex + dy). 

If the two factors {ax + by) and (c# + dy) are multiplied 
together, the product is found to be 

acx 2 + (be + ad)xy + bdy 2 . 

By comparing this product with the trinomial 

Ax 2 + Bxy + Cy 2 , 

we note that it is necessary to find four numbers a, b, c, d, such that 

ac = A, be + ad = B } and 6d = C. 



Sec. 1-21 Introductory Topics 29 

The following example illustrates a trial-and-error procedure, 
which often involves several steps of inspection and testing by 
multiplication ; yet it is a method commonly employed in practical 
work and is recommended here. 

Example 1-23. Factor 6x 2 + llxy - 10y 2 . 

Solution: Here we wish to find a, 6, c, and d which satisfy the identity 

(ax + by) (ex + dy) = acx 2 + (ad + bc)xy + bdy 2 = 6x 2 + llxy - 10?/ 2 . 
Since ac = 6 and bd = — 10, obviously a and c have like signs, and b and d have 
opposite signs. Possible values for a and c are ± 1, ± 2, =fc 3, and =fc 6. Possible 
values for 6 and d are db 1, =fc 2, ± 5, and =fc 10. By trial and error we find the 
correct selection to be a = 2, c = 3, b = 5, and d = - 2. This selection meets the 
requirement because 

(2* + 5y) (3x - 2y) = 6z 2 + llxy — I0y 2 . 

Type 6: Factoring by Grouping. An expression which does not 
fall directly into one of the given type forms can sometimes be 
reduced to one of these forms by a suitable grouping of terms. 
The following examples illustrate the procedure. 

Example 1-24. Factor Sax - 5bx + Gay - 1(%. 

Solution: First, group within parentheses the terms having a common factor. 
Thus, 

Sax - 5bx + Say - IQby = (Sax - 5bx) + (6a?/ - 1(%). 

Then, in each group, factor out the common quantity. In this case, 

(Sax - 5bx) + (Sax - 1(%) = x(Sa - 5b) + 2y(Sa - 56). 
Finally, factor out the quantity common to the terms obtained. (This quantity 
will often be a multinomial factor.) The result is 

x(Sa - 56) + 2y(Sa - 56) = (3a - 56) (x 4- 2y). 
An alternate method of grouping would give us the following results : 
First, 

3ax - 5bx + Say - lOby = (Sax + Say) + (- 56s - 1(%). 
Then, 

(Sax 4- Say) 4- (- 56s - 1(%) = 3a(x + 2y) - 56(z + 2y). 
Finally, 

3a(z + 2^/) - 5b(x + 2y) = (x + 2y) (3a - 56). 

Example 1-25. Factor z 2 - ?y 2 4- 62/0 - 9* 2 . 

Solution: By grouping the last three terms, we may rewrite the given expression 
as the difference of two; squares. Thus, 

x 2 - y 2 + Syz - 9z 2 = x 2 - (y 2 - 6^ + 9s 2 ) 
= z 2 - (2/ - 3*) 2 
= [* - (y - 3*)1 [* + (y - 3«)1 
= (z - 2/ 4- Sz) (x + y ~ 3*). 



30 Introductory Topics Sec. 1-21 

Example 1-26. Factor 4s 4 + 3x V + s/ 8 . 

Solution: By adding and subtracting »V, we may rewrite the given expression 
as the difference of two squares. Thus, 

4 X 4 + 3^4 + y s = 4s 4 + 4s 2 ?/ 4 + </ 8 - * 2 2/ 4 
= (2s 2 + 2/ 4 ) 2 - xW 
= [(2s 2 + t/ 4 ) - xy*] [(2s 2 + 2/ 4 ) + xy*] 
= (2s 2 4- y 4 - Z2/ 2 ) (2s 2 + 2/ 4 + si/ 2 ). 

EXERCISE 1-5 

Factor each of the following: 

1. 3s + 62/. 2. 6s 2 + 4s2/. 3. 4s + 14. 

4. 2a - db. 5. 3as + 2a. 6. 4as 2 - 2/a 2 s 3 . 

7 - as -f 2cs - s 2 . 8. 2a* - 360* + 6c* 3 . 9. ax - 2ay + 3a*. 

lol 3a6 + 9ac - 66c. 11. W + 3y» - at/ 2 . 12. s 3 - 5s 2 2/ + 6s?/. 

13 a; 2 - 16. 14. y 2 - 64. 15. 4s 2 - 9. 

16. 16s 4 - 25y*. 17. 49s 2 - 121. 18. 25a 2 - 166 2 . 
19. 9y* - a 2 . 
22. 4a 2 s 4 - 81. 

25. 49s 2 ?/ 2 - 144a 2 6 2 . 26. a 2 - ±- 27. 36s - s* 

28. 4s 2 
31. 



• 20. 1 - s 2 ?/ 2 . 21. a 2 s 4 - 9a 4 s 2 . 

23. 0.01 - & 2 . 24. - a^x 2 - ax*. 



- 16s 4 . 29. 2s 3 + 16. 30. 3s 3 - 81?/ 



3 



32. a» - 2166». 33. 125pV + r 16 . 

34 a' + 6". 35. x« - 2/> 2 . 36. 2»V - 128^. 

37*. Six" - 3(22/ + &)». 38, (a + 26)3 +.(3 C ^4d)'.__3?^( ? _ i ^- (2- ^) 3 - 
"> ^ „.,' " '-"41. 216«» -l/< 42 - * 2 - "* + 3b - 

~^^ 12x3 + *.. 44. 8x2 + 2x - 15. 45. xV - l&W + 81. 

- 6 47. x 4 - 8x^ + 15. 48. x* - 11* + 30. 

£T-~2x - 3. . 50. 2x* - 3x + 1. 51. x* - 4x</ + 4tf*. 

„. x" - 2* 2 + 1. 53. x* - x - 12. • 54. x* - 3x - 10. 

55! x* + 3*2 + 2. 56. *• + 4x3 + 4. 57. 1 + 30x3 + 225 xe. 

58 x< - 10x= + 9. -5Q. 5x3 + 10x3 _ 40x. 60. a' + ba - 24. 

6l! 2x2 _ nx _ 6. 62. x2 + 16x + 64. 63. 2x2 + 5x - 3 

64. 18x2 + i5x - 25. 65. 6x2 _ 3 7x + 6. 66. 4x2 + 32 x + 15. 

67 x2 - 1 2x + 0.36. 68. 4<te*tf + 35xyz - 1&*. 69. ax + 3x + 2ay + Qy. 
70. x3 + 3x2 _ 7 x - 21. 71. 8x» _ IV - lto + 15. 

72. ax + 26* - ex + 3ay + 66*/ - 3cy. 73. 2ax + 3y - xz + 6a + xy - 3z. 




74. xa6 - xyz - 2a6j/ + 2y 2 z. 



75. x 3 + x 2 - 3x - 3. 



1-22. GREATEST COMMON DIVISOR 

A common divisor of several polynomials (or integers) is a poly- 
nomial (or integer) which divides each of them. For example, 
2 and 3 are common divisors of 12 and 18. Also, x and x + y are 
common divisors of x* (z 2 - y") and x 3 (x 2 + 2xy + y>). 

A greatest common divisor (G.C.D.), also known as a highest 
common factor (H.C.F.), of two or more polynomials (or integers) 



Sec. 1-22 Introductory Topics 31 

is a polynomial (or integer) with the following two properties: 
It is a common divisor of the given polynomials (or integers) ; 
also it is a multiple of every other common divisor of the given 
polynomials (or integers). 

It follows that, for integers, a G.C.D. is a common divisor of 
greatest absolute value. It also follows that, for polynomials, a 
G.C.D. is a common divisor of highest degree. 

For example, a G.C.D. of 12 and 18 is +6 or -6, since ±1, ±2, 
±3, and ±6 are the only common divisors, and 6 and —6 are those 
of maximum absolute value. This example indicates that a set of 
non-zero integers will have two greatest common divisors, d and 
— d, one being positive and the other negative. 

For polynomials with real coefficients, if d is a G.C.D., then a • d 
is also a G.C.D. for every real number a not equal to 0. It follows 
that infinitely many greatest common divisors exist. However, for 
polynomials with integral coefficients, a G.C.D k should be a poly- 
nomial of the same type. Example 1-27, which follows, illustrates 
the fact that in this case a G.C.D. is uniquely determined except 
for sign. 

Since a G.C.D. of given polynomials divides all of them, it must 
contain as a factor each of the distinct prime factors occurring as 
a common factor of all the given polynomials. Since, however, a 
G.C.D. must divide any common factor of the polynomials, it must 
contain each of the distinct common primes raised to the highest 
common power. Thus, a G.C.D. of x 3 (x + y) 2 and x* (x — y) (x + y) 
must contain the prime factor x to the third power, that is, 
to the highest common power. Similarly, it rftust contain (x + y) 
to the first power. Thus, x*(x + y) is a G.C.D., since no further 
common prime factors occur. 

When no common prime factors occur, a G.C.D. is 1. 

Example 1-27. Find a G.C.D. of 3x*y*(x* - 4s/ 2 ) and Sxy 2 (x 2 - 4xy -f 4z/ 2 ). 

Solution: We shall begin by writing each of the expressions as the product of its 
prime factors, as follows: 

Zxhjs(x* - ±y*) = (3) (x) (x) (y) (y) (y) (x -f 2y) (x - 2y), 
and 

6xy*(x* - ixy + ±y 2 ) = (2) (3) (x) (y) (y) (x - 2y) (x - 2y). 

The different prime factors are 2, 3, x, y, (x -f 2y), and (x - 2y), of which only 
3, x } y } and x - 2y are common to both polynomials. We now form the product of 
these common factors, using for each the maximum common power. Hence, a 
G.C.D. is 

3xy 2 (x - 2y). 



32 Introductory Topics Sec. 1-23 

1-23. LEAST COMMON MULTIPLE 

A common multiple of two or more polynomials (or integers) 
is one containing each of the given ones as a factor. Thus, 36 is 
a common multiple of 6 and 9, and x 2 — y 2 is a common multiple of 
x — y and x + y. 

A least common multiple (L.C.M.) of two or more polynomials 
(or integers) is a polynomial (or integer) with the following 
properties: It is a common multiple of the given polynomials (or 
integers) ; also it is a divisor of every other common multiple of 
the given polynomials (or integers). 

It follows that, for integers, an L.C.M. is a common multiple of 
least absolute value. It also follows that, for polynomials, an 
L.C.M. is a common multiple of lowest degree. 

For example, an L.C.M. of 6 and -8 is 24 or -24, since the only 
common multiples are ±24, ±48, ±72, • • • , and 24 and -24 are 
those of minimum absolute value. This example indicates that a 
set of non-zero integers will have two least common multiples, m 
and — m, one being positive and the other negative. 

For polynomials with real coefficients, if m is an L.C.M., then 
a • m is also an L.C.M. for every real number a not equal to 0. 
It follows that infinitely many least common multiples exist. How- 
ever, for polynomials with integral coefficients, an L.C.M. should be 
a polynomial of the same type. Example 1-28, which follows, illus- 
trates the fact that in this case an L.C.M. is uniquely determined 
except for sign. 

Since an L.C.M. of given polynomials is a multiple of all of them, 
it must contain as a factor each of the distinct prime factors occur- 
ring as a factor of any one of the given polynomials. Since, how- 
ever, an L.C.M. must be a multiple of any common multiple of the 
given polynomials, it must contain each of the various distinct 
primes to the highest power occurring anywhere. For example, an 
L.C.M. of x s (x + y) 2 and x*(x - y) (x 4- y) must contain the vari- 
ous distinct prime factors, which are x, x + y, and x - y. For x, 
the highest power occurring anywhere is the fifth power ; f or x + y, 
the highest power is the second ; for x — y, the highest power is the 
first. An L.C.M. is therefore x 5 (x + y) 2 (x-y). 

Example 1-28. Find an L.C.M. of 4x2 - ± Xy § X 2 _ 6, and 9x 2 - 18x + 9. 

Solution: We shall first rewrite each of the expressions in factored form. Thus, 
4x2 - 4x = (2) (2) (x) (x — 1) = 2 8 x(x - 1), 
6x2 - 6 = (2) (3) (x + 1) (x - 1), 
and 

9x2 _ 18x + 9 - ( 3 ) ( 3 ) ( x - i) (s - i) = 32(x - 1)2. 



Sec. 1-24 Introductory Topics 33 

The distinct prime factors are 2, 3, x, x + 1, and x - 1. The greatest powers for 
these are 2, 2, 1, 1, and 2, respectively. An L.C.M. is therefore 

2 2 • 3 2 • x • (x + 1) • (3 - l) 2 . 



EXERCISE 1-6 

In each problem from 1 to 12, find a G.C.D. of the given expressions. 
1. 4, 14, 36. 2. 9, 21, 33. 3. 4, 7, 39. 

4. x + y, x 2 - 2/ 2 . 5. x 3 - 2/ 3 , x - y. 6. 3a6, 12a 3 6, 6a 3 6 2 . 

7. 9x 3 */ 2 , 15x 4 */ 2 , 21x 6 i/. 8. x - 3, x 2 - 9, (x - 3) 2 . 9. ±x*y% 8xij 7 z*, Ux s y*z 2 . 

10. x + 2, x 2 - 4, x 3 + 2x 2 - 4x - 8. 

11. x 3 - x 2 - 42x, x 4 - 49x 2 , x 2 - 36. 

12. x 4 + 2x 3 - 3x 2 , 2x 6 - 5x 4 + 3x 3 , x 3 + 3x 2 - x - 3. 

In each problem from 13 to 22, find an L.C.M. of the given expressions. 
13. 6, 8, 12. 14. 8, 45, 54. 

15. xy, 6x2, 8yz. 16. 4x 2 , 5x 4 , 20x. 

17. 4x 2 i/ 4 z 5 , 9x 6 i/ 2 2 3 > 6x 4 yz*. 18. 2a + 4, a - 3, a 2 - 9. 

19. x - 2, x + 2, x 2 - 4. 20. 2x + 8, 3x - 6, x 2 + 2x - 8. 

21. (x 2 - 49) (x 3 - 8), (x - 7) (x + 7) (x - 2) (x 2 - 4), (x - 3) (x - 2). 

22. 2x 4 - 22/ 4 , 6x 2 + \2xy + 6y 2 , 9x 3 + 9yK 

1-24. REDUCTION OF FRACTIONS 

From (1-24) under operations with fractions, it follows that a 
fraction a/b, in which b ¥=0 t is not changed if both the numera- 
tor and the denominator are multiplied or divided by the same 
quantity, provided that the quantity is not zero. That is, if k ¥= 0, 

n nn\ a - a / k 

(1 ~ 50) b~W 

For example, i _ 2^ _ 2 H * - ^ = ? 

2 "" 2 • 2 " 4 ' 6 " 6/2 3 ' 

The fundamental principles of (1-49) and (1-50) are applied in 
reducing a fraction to lowest terms and in changing two or more 
fractions with different denominators into equivalent fractions with 
a common denominator. 

The reduction of a fraction to lowest terms, that is, to a form in 
which all common non-zero factors are removed from both the 
numerator and the denominator, is accomplished as follows: 

First, factor both the numerator and the denominator into 
prime factors. 

Then, divide both the numerator and the denominator by all 
their common factors. 



34 Infroductory Topics Sec. 1-24 

It should be noted that this reduction can also be accomplished 
by dividing the numerator and the denominator by their highest 
common factor. 

The following examples will illustrate the reduction of fractions 
to lowest terms. 

30 
Example 1-29. Reduce jt to lowest terms. 

Solution: Factoring the numerator and the denominator and dividing both by 
their common factors, we have 

30 _ 2 ; 3 • 5 5 

42 2 • 3 • 7 7 ' 
The common factors of the numerator and the denominator are 2 and 3. 

Example 1-30. Reduce tr-f- to lowest terms. 
2x 2 y 

Solution: Factoring into prime factors and dividing out common factors, we have 

2 • 3 ; x * y • y _ 3y # 

2 • x • x • y x 

In this fraction the common factors are 2, x, and y. 

Example 1-31. Reduce -z-z ^ ^-r-r to lowest terms. 

6x 2 - 3xy - ISy 2 

Solution: We have 

6s 3 - 24xy 2 2 * 3x(x - 2y) (x + 2y) = 2x(x -f 2y) , 

6z 2 - 3sy ~ ISy 2 ~ 3(s - 2y) {2x + 3y) "" 2x 4- 3y 

It is important to note that in a fraction of the type a ~T - > where 

the numerator and the denominator have a common term, any 
attempt to simplify the fraction by cancelling out this common 

term can lead only to an absurdity. For, quite obviously, a y 

.. . a -j- x 

does not equal either "T - or - unless a = 1 or 0. For example, if 

1 -f- X X o I o 

we attempt one of these simplifications with the fraction > 

5 4 5 3 ^" 
we reach the obvious contradiction - = - or - = - • To avoid this 

6 5 6 4 

common error, it is important to remember that only common 
factors, not common terms, may be cancelled. One should make 
certain that the common quantity is a factor of the entire numera- 
tor and of the entire denominator. 

1-25. SIGNS ASSOCIATED WITH FRACTIONS 

It follows from (1-49) that we can multiply both the numerator 
and the denominator of a fraction by -1 without changing the 



Sec. 1-25 Introductory Topics 35 

value of the fraction. However, if just one of these is multiplied 
by — 1, then the sign of the fraction must be changed in order to 
keep its value unaltered. Thus, in effect, a minus sign before a 
fraction can be moved to either the numerator or the denominator 
without altering the value of the fraction. We have, as previously- 
stated in (1-22), 

n __ — n _ n 
, d ~~ d ~~ — d 

and i-ii i 



x - y x - y - (x - y) y - x 

We see, then, that any two of the three signs associated with a 
fraction, namely, the signs of the numerator, the denominator, and 
the fraction, can be changed without changing the value of the 
fraction. In general, the rules for changing signs in a fraction are 
as follows: 

Changing the signs in an even number of factors in the numera- 
tor or in the denominator, or in both, does not change the sign of 
the fraction. 

Changing the signs in an odd number of factors in the numerator 
or in the denominator, or in both, does change the sign of the 
fraction. 

2x — 2x ( ) 
Example 1-32. Find the missing quantities in — = - -, r = -^-r • 

,-j 7 . . ZlX ZlX £X 

Solution: — = — = ~ • 

o o — o 

Example 1-33. Change the fraction — — to an equivalent fraction with 
denominator x — y. 

Solution: Since x - y = - (y - x), we make the following changes in signs: 
a ___ - a __ - a 
V - x ~~-(y-x)~x-y' 
a a a 



y - x - (y - x) x - y 

EXERCISE 1-7 

1. Find the missing quantity in each of the following equalities: 

a * 7 ~ 21 15 5 2 8o 

d ' te» ~ ( ) ' C ' Sy ~ ( ) 2a;+a te+ 2a» ' 

2x -2x . x +y _ ( ) > . x -a _ ( ) # 

g ' I - x 2 (1 + x) ( ) ' ' x -y y -x ' x - b 6» - x* 



36 Introductory Topics Sec. 1-25 

2. Reduce each of the following fractions to lowest terms. 



102010 . 8a 2 b 3 6s 4 y 8 

350470 * 12a<6 ' C# 20* V ' 

36x 4 2/*s 3 48x 5 ?/ 8 f x A - 6x 2 

96x 3 y 7 z 2 * e# 24xV 4- 40x 4 ?/ 5 ' 3x 3 - 18x " 

a 4-26 . x 2 y* - x 3 y 2 . 9x 2 - 49 

4a 2 + 8a6 * a; - y U 3x 2 + 13x + 14 ' 

2^2 — 7a; - 15 , 6a: 2 - 5x - 6 , x* - 81 

k. 



J ' x 2 - 25 4x 2 + lOx - 24 x 6 + 729 

X 2 z (y „ 3)2 fl a _ 86 3 a; 3 - 27y 3 

m - 4 — (x — 3/) 2 ' "' a 2 - 4& 2 # °* a: 2 + 3sy + 9y 2 ' 

a; 2 - (a 4- 6)x + <?& (x + y) (x - y) (y - z) , 

P * [(2a: 4- a) 2 - (* + 2a) 2 ] (a: 2 - 6) ' q * (?/ 2 - a; 2 ) (z - ?/) 

18x 2 4- Sxy - 10y 2 t 9a 2 + 6a6 4- & 2 . 

f# 21x 2 - 26x2/ 4- 8y 2 ' S * 6a 2 - a& - 6 2 * 

. x 2 +xy x 2 4- 2xy 



a: 3 4- a: 2 y 4- a^ 2 4- y 3 ' x 2 + x + 2y - 4a/ 2 

1-26. ADDITION AND SUBTRACTION OF FRACTIONS 

The sum of two fractions with the same denominator is the 
fraction whose numerator is the sum of the numerators and whose 
denominator is the given common denominator. That is, in (1-25), 

(1-25) S + h^r*' 

For example, x + 2y 

3 + 3 3 

To add fractions with different denominators, first change the 
fractions to equivalent fractions having a common denominator, 
and then write the sum of the new numerators over this common 
denominator. 

Ordinarily, the common denominator which is chosen is a least 
common multiple of the given denominators, since this leaves the 
fewest possible common factors in the numerator and denominator 
of the resultant fractions. However, the same result is obtained, 
after concellation of all common factors, no matter what common 
denominator is used. The method of finding an L.C.M. was devel- 
oped in Section 1-23. 

The difference of two fractions, ~ — ^ > has been defined in 

o a 

Section 1-2. Thus, by (1-10), when neither b nor d is zero, 

< l -»> f-5-f+C-a)- 



Sec. 1-26 Introductory Topics 37 

Applying (1-22), we have 

C" 52 ' 5-5 = 5 + ^- 

The following procedure is suggested for adding (or subtracting) 
fractions. 

1. Find a least common multiple of the given denominators. 

2. Change each fraction to an equivalent fraction having the 
L.C.M. as the denominator in the following way: For each 
fraction, note which factors are in the L.C.M. but not in the 
denominator of the given fraction. These factors may be found 
by dividing the L.C.M. by the denominator of the given fraction. 
Then multiply the numerator and the denominator of the given 
fraction by these factors. 

3. Write the sum (or difference) of the numerators of the 
new fractions found in step 2 over the L.C.M., and reduce the 
resulting fraction to lowest terms. 

The following examples will indicate a procedure which should 
be followed until some skill in working problems has been attained. 

3 5 7 
Example 1-35. Express j - -r -f - as a single fraction reduced to lowest terms. 

Solution: Step 1. The methods in Section 1-23 give us 36 as the L.C.M. of 
the denominators. 

Step 2. Divide the L.C.M. successively by the denominators 4, 6, and 9 to get 

36/4 = 9, 36/6 = 6, and 36/9 = 4. 
Change the given fractions to fractions having 36 as the denominator in the 
following way: 

3i?_27 5^6 = 30 7 - 4 28 
4 • 9 ~ 36 ' 6-6 36 ' 9-4 36 ' 

Step 3. Combine the new fractions to obtain 

3_5,7_27_3028 27 - 30 + 28 _25 # 
4 6 + 9 ~ 36 36 + 36 36 36 ' 

Note that adding (or subtracting) the numerators and the denominators of the 
given fractions leads to absurdities. Thus, it is not true that = + o = * * Also, 

dropping the common denominator leads to absurdities. Thus, it is not true thit 

I + I-- : 

3x 2 5x — 2 
Example 1-36. Express — 7-3 - -5 i i as a sm gl e simplified fraction. 

2 2 

Solution: Step 1. Since x 2 - 4 = (x + 2) (x — 2) and — ^ = 5 > an 



38 Introductory Topics Sec. 1-26 

L.C.M. of the denominators is x 2 - 4. The given expression may be written 
as follows: 

3x 2 5x - 2 



x + 2 x - 2 x 2 - 4 

Step 2. When the L.C.M. is divided by the denominators fa + 2), fa — 2), and 
(x 2 — 4), the results are 

x 2 - 4 _ x 2 - 4 __ a: 2 - 4 _ , 

¥+T - * - * ^T ~ x + ^ o^l - x - 

Multiplying the numerators and the denominators of the given fractions by these 
factors, we obtain 

3x * fa - 2) _ 3x 2 - 6x 
(x + 2) • fa - 2) ~ x 2 - 4 ' 

2 • fa + 2) 2x +4 

(s - 2) • (a + 2) ~ x 2 - 4 ' 
5s - 2 _ 5x - 2 _ 
x 2 - 4 "~ x 2 - 4 ' 

Step 3. Therefore, the desired result is obtained in the following way: 
3x 2 5x - 2 _ (3x 2 - 6x) + (2x 4- 4) - (5x - 2) 



a? + 2 ' s - 2 a: 2 -4 x 2 - 4 

3x 2 - 9x 4- 6 3 fa - 1) (a; - 2) __ 3fa - 1) # 
~ x 2 - 4 "" x 2 - 4 x 4- 2 ' 



EXERCISE 1-8 

Perform each of the indicated operations and express the answer in lowest terms. 

4 1 + 5 - A 

8 + 6 12 
ft 5 17 ,19 



1. 


1 2 7 

2 3 + 8* 




3. 


3-5+1. 
7 5+ 21 




5. 


3+ 20 


5 
10 ' 


7. 


1 1 




1 + X x — 


1 


9. 


3x - 2 x - 

x z - 1 'l- 


-4 
- 1 


11. 
is 


1 £ _ 6 bi 

a: x 2 x 3 ' 

1 2 





+ ■ 



8. 


2 13 


1 - x 1 4- z 1 - z 2 


10. 


2x Sx 2 
x 4- 2 x - 1 + *' 


12. 


1 1 - * - 
x -1 ' - 1 -X 


14. 


x - 2 x + 2 8x 
x + 2 2 - x 4 - x 2 


!, 


x 4-3 



3-1 X 2 - 1 ^fa - l) 2 

x -2 x4-2 

x 2 + 5x 4- 6 "*" x 2 4- 7x 4- 12 ^ x 2 + 6x 4- 8 

M. » - 2 +1. 17.1+. + - 2 * 



x 2 — 3x x — 3 x 1 — x 

18 g 4- g + 1 - 2x * ~" 3 . 19 2x Sx 



x + 1 x x 2 + x x 2 - y 2 x 2 - 2xy + y 2 



Sec. 1-27 Introductory Topics 39 

3 1 



20, 



4 - x 2 x - 2 



22.-JL..+4+: 6 * 



x - 1 2 - x 

24. _* y —+ xy ■ 

# - 2/ & - y x 2 - y 2 

Sx 2x 



26. 



OJ a -f 6 a ~ 6 

a — 6 a -f- 6 


2ab 
a 2 - b 2 


23.i+? + A. 

x y xy 




25. 1 + ' 

x + y x - y 


xy 


x 2 - y 2 


4xy 





x 2 — y 2 x 2 — 2xy + y 2 x 3 — x 2 y — xy 2 -f y 3 



1-27. MULTIPLICATION AND DIVISION OF FRACTIONS 

The product of two fractions is the fraction whose numerator is 
the product of the numerators and whose denominator is the 
product of the denominators of the given fractions. That is, by 
(1-29), if neither 6 nor d is 0, 
, oqn a c _ ac 

(1 ~ 29) Vd'bd 9 

To illustrate, 

2 6 _ 2j_6 _ 4 , x - 1 x + 3 _ x 2 + 2x - 3 

3 ' 7 ~ 3 ■ 7 ~~ 7 ' and z + 2 ' z - 2 ~ x 2 - 4 

A special case of (1-29) is worth noting. To multiply a fraction 
by a number or expression, we multiply the numerator by that 
number or expression. Thus, 

b _ a b __ a • b _ ab 

c ~" I c "" 1 • c ~~ c 

For example, , 1N 

2 • - = - > and (x — 1) • - = — • 

5 5 J x x 

The quotient of two fractions, T / -. > has been defined in 

bi a 

Section 1-2 and evaluated in Section 1-7. Thus, if no factor in a 

denominator is zero, 

/, Q m a /c a d ad 

(1 ~ 30) &/d = 6'c = te - 

To illustrate, 

5 /3 __ 5 2 _ 5j_2 _ 10 . 3x 2 /tef __ 3z 2 j£_ _ 3a; 2 * y 7 _ j£_ 

7/ 2 ~ 7 ' 3 " 7 • 3 "" 21 ' y 3 I y 7 " y z ' 6x 5 y z • 6z 5 2x*' 

In practice, we divide out all factors common to the numerator 
and denominator before proceeding with the actual multiplication. 
If necessary, the numerator and the denominator of each given 
fraction should be factored. 

%2 2x 2x 2 Sx y 

Example 1-37. Multiply . — —= by - s and simplify the result. 



40 Introductory Topics Sec. 1-27 

Solution: The work may be indicated as follows: 

x 2 - 2x t 2x* - 3s - 9 _ x(x - 2) (x - 3) (2s 4- 3) 
2x* + 5s + 3 ' s 2 - 9 ~ (s 4- 1) (2s 4- 3) (x + 3) (a; - 3) 

__ j;(3; - 2) __ x 2 - 2s 
"" (s 4- 1) (s 4- 3) ~~ x 2 4- 4s 4- 3 ' 

w i i oo 7^- -j ^ 2 - 5s + 4 , s 3 - 4s 2 -f s - 4 
Example 1-38. Divide^— 3^-^ by ^-^ 

Solution: By (1-30), 
s 2 - 5s + 4 s 3 - 4s 2 4- s - 4 __ (s 2 - 5s -f 4) • (2s - 1) 



s 2 - 3s - 4 ' 2s - 1 (s 2 - 3s - 4) • (s 3 - 4s 2 4- s - 4) 

- (s - 1) (s - 4) (2s - 1) 
~ (s + 1) (s -4) (s -4) (s 2 +l) 
(s - 1) (2s - 1) 



(s + 1) (x - 4) (s 2 + 1) 

1-28. COMPLEX FRACTIONS 

The fractions we have been discussing so far may be called 
simple fractions, to distinguish them from the fractions which we 
now discuss. 

A fraction which contains other fractions in the numerator or 
in the denominator is called a complex fraction. Since the simpli- 
fication of a complex fraction is essentially a problem in division, 
we first reduce the numerator and the denominator of the complex 
fraction to simple fractions and then proceed as in division. 

Example 1-39. Simplify t r • 

s y 

First Solution: The numerator of the given complex fraction reduces to the simple 

x -4- v V 4~ X 

fraction > and the denominator reduces to • Hence, we have 

s , xy 

s -f y 

x __ (s + y) • xy = 



y+g x • (?/ 4- s) 

st/ 

Alternate Solution: Frequently it may be more convenient to multiply both the 

numerator and the denominator of the complex fraction by an L.C.M. of the 

denominators of all simple fractions occurring in the given complex fraction. In this 

y 1 1 

example, the simple fractions are - > - > and - > and an L.C.M. of their denomi- 

i. • xx y 

nators is xy. 

Therefore, by multiplying the numerator and the denominator of the complex 

fraction by xy } we get 

y\ 
xy 



(■+!)■ 



xy 



xy +y 2 = (sj- y)y 



Sec. 1-28 Introductory Topics 41 

EXERCISE 1-9 

In each of the problems from 1 to 20, perform the indicated operations and 
express the answer as a simple fraction in lowest terms. 
, 3 ]8 2 5 # 

' 5 ' 155 ' 3 7 

. 3 | 4 iy tofi_ % 

7 ' 6t/ ' 9x ' 16?y 3 # 



5 



14 /28 fi Jx_ / 16x 3 g 

" 9/45* 17t/ 3 / 68?y 6 ?/ 

6a 2 6 3 3a 3 6 - 27a; 3 ?/ / 15^ 

15a 3 6 5 ' 48a6 2 ' 72r?/ 3 / 16x ' 

15a 4 6 4 / 54aft 2 3af J02_ Jty^ 

15a 3 / 14a 3 6 2 ' 4?/ ' 27x 3 ' 15z 3 ' 

n 9a; 4 ?/ 3 a: 2 - 6x + 5 12 x 3 - 8 a 2 + 2s - 3 

x 2 - 25 ' 6a; 3 ?/ 2 ' * 9 - x 2 ' x 2 + 2a; + 4 ' 

x 2 + 2x - 3 x 2 + x - 6 x 4 + 27a; . x 3 ± 3a; 2 - 2a; - 6 

(a; - 7) 2 : a; 2 - 5ar - 14 ' a; 4 - 4x 2 ' x(x 2 - 5a; + 6) 

27a; 3 - 8 4x 2 - 25 



15. 
16. 
17. 
18. 
19. 
20. 



6a; 2 + 19a; + 10 9x 2 - 12a: + 4 
6a; 2 + 5a; - 21 6a; 4 1- 36a; 3 2x 2 - 5x - 12 
4a; 2 - 9a; - 28 ' 9 - 4a; 2 ' 9a; 3 + 21x 2 
2x 2 ± 5x + 3 # 3a; 2 - 20a; + 12 ^ 6a; 2 + 5a; - 6 # 
5a; 2 - 24a; - 5 ' x 2 + 3a: + 2 *' 4a; 2 + 9a; + 2 ' 
x 2 - (2x - 3*) 2 _^ l~ 4x 2 = (3g - a;) 2 9s; 2 - (a; - 2?/) 2 1 
(x - 3z) 2 - 4?/ 2 : L(2?y - x) 2 - 9z 2 ' (3z - 2y) 2 - x 2 J 
8x 2 - 2x - 15 9 - 4xH 4x 2 + 12x + 9 



j" 8x 2 - 2x - 15 ^ 9 - 4x 2 H 
L 4?/ 8 * 9 *' 36?/ 7 * 7 J 



22. ,7 x " • 23. 



4?/ 8 z 9 * 36?/ 7 * 7 J 48x 2 ?/ 3 + 60?/ 3 

(x 4 - 625) (x 2 - 9) 3x 4 -f 75x 2 
(x + 5) 2 (x - 5) 3 : x 2 + lOx + 25 ' 
In each of the problems from 21 to 38, simplify the complex fraction. 

1 _2 3 

2 3 + 4 

5 ^6 ^8 
16x 2 ?y 3 

9x 

18x 3 y 

Axy 

a 

T7^ """«_l + i' 

j/ 2 . 6^6* 3 3a T 3 

Itt.|IfL. 31.5ll£LL. 32.1^2. 

2_2_4 !_ 4 oX 20 

a; 8 a; *» a: 2 - 4 



91 


2 3 
3*4' 


7 
2 




421. 


5 
16 






24. 


7 
18 


4 
15 


-a 


1 
2 


3 
5 


♦s 


27. 


8x 3 - 
2x - 


-2,3 

-y 


■ • 



5 3 
9* 10* 


18 
19 


64 
81 




6x?/ 
5 




4x 2 




2y 

X 


-3 



25. -ihr • 26. 



28. , * „ • 29. 



42 Introductory Topics Sec. 1-28 

1 x , 1 - 2x 



x - 5 + 
x - 3 - 



33. £ + *. 34. 2LZ*. 35. ^±1 



34. 


x • 


-2 - 


z 


— 


2 


x • 


-4 - 


x 


1 


4 




z 


+ 2/ 




a; 


-2/ 


37. 


£ 


-2/ 




z 


+ v 


a; 2 + y 2 




a: 2 


-y 2 



1 -2a; 



# -|- 4 x-4 2a? + 1 a: 

1-s 1 + a; x +y x - y a; - 2j/ fi 2t/ 



1 L 2y - ad 



*fi 1 + x * ~ g 27 LzJ g + y Qft s + 1 L 2z/ 

x + l x - l ' a; 2 +y 2 a: 2 -!/ 2 ' aM-Jj/f 1 

a - 1 a: + 1 x 2 -y 2 x 2 +y 2 x 2 - 1 a: - 1 

Simplify each of the following expressions: 

, 1 1 

a; +- a; 

a; a; 

1 — 1+- a; +- x +- 

39. 2 — j — £. 40 # 2L- 41. \ 



X + r-rr 1 + r 

x + 1 a; - 1 

1111 



" ' * -1 

V 35 s 4- v 8 J4k a? 3 V s X 2 v 2 

42. — i — = — * * ^% • 43. ^ V - ~ *r • 

i _ i a; 2 -?/ 2 JL __ i_ lo-i 

a; 2/ a; 2 y 2 a; y 

*e*i-)(s^-,-k)*(— +^- 

«-(*-^)(»- a ^^)*( 2 -ifT)- 

1-29. LINEAR EQUATIONS 

Introduction. In Section 1-13, an equation was defined as a state- 
ment of equality between two algebraic expressions. In this section 
we shall discuss equations in one unknown of the simplest type, 
called linear equations, typified by the form 
(1-53) , ax-b } 

where a and b are specified numbers and a^O. Some ideas pertain- 
ing to equations in general are needed first. 

Solution or Root of an Equation. We shall use the term unknown 
to designate a literal quantity that appears in an equation and is 
not regarded as specified at the outset. A system of values of the 
unknowns which, when substituted for them, makes the equation a 
true assertion is called a solution of the equation. A solution is 
also called a root when only one unknown is involved. Thus, the 
values x = 2 and y = — 1 define a solution of the equation 
3# + 2# = 4, since the equation is satisfied for these values ; that is, 
3(2) + 2(-l) = 4. 



Sec. 1-30 Introductory Topics 43 

Equivalent Equations. Two equations are equivalent if they have 
exactly the same solutions. For example, 2x + 11 = Ix - 4 and 
Ax + 22 = lAx - 8 are equivalent, since, as will be seen, each has 
exactly one root, namely, x = 3. However, 2x + 11 = Ix - 4 and 
2x 2 + llz = 7z 2 - 4# are not equivalent, because the latter equation 
has, in addition to the root x = 3, the root x = 0. 

Operations on Equations. Each of the following operations on an 
equation yields an equivalent equation : 

Adding the same expression to (or subtracting the same expres- 
sion from) both sides. 

Multiplying (or dividing) both sides by the same non-zero 
number. 

For any solution of F = G also makes F +H=G+H true, and 
is therefore a solution of this latter equation. Conversely, any 
solution of F + H - G + H is one of F = G, because 
F = F + H-H = G + H-H = G. 

Likewise, any solution of F = G is one of aF = aG, provided that 
a is a non-zero number, and conversely. 

Transposition of Terms. Transposing a term of an equation con- 
sists in moving the term from one side of the equation to the other 
and changing its sign. This operation is equivalent to adding the 
same quantity to both sides (or subtracting the same quantity from 
both sides) . For example, consider the equation 

2x+ 11 = 7x - 4. 

If we transpose 2x from the left side to the right, and transpose 
-4 from the right to the left, we obtain 

11 + 4 = Ix - 2a?. 
In effect, we subtracted 2x from each side and added 4 to each side. 

1-30. LINEAR EQUATIONS IN ONE UNKNOWN 

An equation in the form shown in (1-53) is called a linear equa~ 
tion in one unknown. We shall show that such a linear equation 
has one and only one root, namely, 

(1-54) x = - • 

a 

If each side of the equation ax = b is divided by a, the result is 

b 

x = - > 
a 



44 Introductory Topics Sec. 1-30 

which is equivalent to the given equation. Therefore, (1-53) 
clearly has one and only one solution, namely, b/a. 

It should be noted that if we allow a to be in (1-53), there are 
two possibilities. Either no solution exists, when 6^0, since for 
no number x is it true that a*x = 0'X = = b¥=0; or else every 
number # is a solution, when 6 = 0, since a*# = 0*# = = & for 
all values of x. 

An equation which is not apparently linear may frequently be 
solved by the theory of linear equations, by replacing the given 
equation by a linear equation to which it is equivalent. The follow- 
ing steps serve as a guide to the method to be used when there is 
only one unknown in the equation. 

1. Clear the equation of any fractions with numerical denomina- 
tors by multiplying both sides of the equation by a least common 
multiple of the denominators of those fractions. 

2. Transpose all terms containing the unknown to one side and 
all other terms to the other side. We may collect the terms contain- 
ing the unknown on either side. 

3. Combine like terms. If the equation now assumes the form 
in (1-53), it is a linear equation and can be solved by dividing both 
sides by the coefficient of the unknown. 

4. Check the result obtained in step 3 by substituting in the 
original equation. While it is desirable to include step 4 to show 
that the number x found in step 3 is actually a solution of the 
equation, step 4 is not a necessary part of the solution process, 
since the operations performed in the preceding steps always 
yield equivalent equations. The purpose of step 4 is to help 
to make certain that there has been no error. 

In step 2, we generally transpose terms containing the unknown 
to whichever side makes solving the equation easier. 

The following explanations should be noted carefully. 

Students at times "transpose" coefficients of the unknown. Thus, 

2x = 6 takes the erroneous form x = 6 - 2 by "transposing" 2 to 

the right side. The correct procedure is to remove the coefficient 

2 by division, since it is a multiplier of x. Therefore, we should 

divide both sides of the equation 2x = 6 by 2 and have 

2x 6 

2" = 2' 
or 

x = 3. 

Errors of this type may be avoided if the student applies the rules 
of algebra properly and checks his solutions carefully. 



Sec. 1—30 Introductory Topics 45 

Multiplying or dividing both sides of an equation by a poly- 
nomial involving the unknown will not necessarily yield an equiva- 
lent equation. When the operation is multiplication, the new equa- 
tion thus obtained may have roots in addition to the roots of the 
original equation. These extra roots are called extraneous roots. 
We then say that the equation is redundant with respect to the 
original equation. When both sides of an equation are divided by 
a polynomial involving the unknown, the new equation may lack 
some of the original roots. It is then said to be defective with 
respect to the original equation. 

If both sides of an equation are multiplied by a polynomial 
involving the unknown, the check in step 4 of the recommended 
procedure is a necessary step in the solution process. An extra- 
neous root can thus be identified. Dividing both sides of an equa- 
tion by a polynomial involving the unknown is not a permissible 
procedure, since roots that are lost cannot be regained. 

It should be noted that a non-linear equation may some- 
times be treated so as to obtain a linear equation which is possibly 
redundant. For example, the fractions in the equations of Problems 
39 to 48 of Exercise 1-10 may be eliminated by multiplying both 
sides of each equation by a least common multiple of the denomina- 
tors of the fractions in that equation. Since this multiplier con- 
tains the unknown, the new equation may have solutions which are 
not solutions of the given equation. Hence, the solutions must be 
checked to see whether or not they actually are solutions of the 
given equation. 

2^ x + 7 
Example 1-40. Solve the equation — = 4 — — 

Solution: We clear the equation of fractions by multiplying both sides by 15, 
which is an L.C.M. of the denominators of the fractions, and obtain 

10* = 60 - 3x - 21. 

Collecting the terms containing x on the left side and all other terms on the 

right, we have 10* + 3* = 60 - 21. 

Combining like terms, we obtain 

13x = 39. 

Finally, dividing both sides by 13, we have 

' x = 3. 

To check, substitute 3 for x in the original equation. The result is 

2l2 = 4 - L+Z , or 2=4-2, or 2=2. 
Therefore, 3 is the root. 



46 Introductory Topics Sec. 1-30 

Example 1-41. Solve the equation 6x — 3y — 1 = by + 2x + 11 for y in terms 
of x. In this case, regard x as specified. 

Solution: Collect the terms in the unknown y ) on the right side, and collect all 
other terms on the left. The result is 

6x - 2x - 1 - 11 = 5y + 3y. 

Combining like terms, we have 

4z - 12 = %. 

Now we divide both sides by the coefficient of the unknown to obtain 

4x - i2 

— — = »• 

Then ' 4* -12 x-3 

V =— — =-2— 

x — 3 

To check, substitute — ~ — for t/ in the original equation, and obtain 

This equation reduces to 

12x - 3(x - 3) - 2 = 5(x - 3) + 4z + 22, 
which becomes 

12z - 3a; + 9 - 2 = 5x - 15 + Ax + 22, 
or 

9x + 7 = 9x + 7. 

x — 3 
Since this result is an identity, y = — — is the solution of the original 

equation, regardless of the value of x. 

Example 1-42. Find two consecutive integers such that four times the first is 
equal to six times the second diminished by 20. 

Solution: Let x be the smaller integer, and x -f 1 the next larger integer. Then, 
from the statement of the problem, we have 

4x = 6(x 4- 1) - 20. 

From this equation, we obtain 

4x = Qx + 6 - 20. 
Then, 

14 = 2x, or x = 7. 

Hence z = 7 and x 4- 1 = 8 are the two consecutive integers. The student 
should carefully check these values by substitution in the original statement of 
the problem. 

Example 1-43. The speed of an airplane in still air is 400 miles per hour. If it 
requires 20 minutes longer to fly from A to B against a wind of 50 miles per hour 
than it does to fly from B to A with the wind, what is the distance from A to B1 



Sec. 1-30 Introductory Topics 47 

Solution: Let x be the distance from A to B. From the data, the speed of the 
plane against the wind, in miles per hour, is 400 — 50 = 350, and the speed of the 
plane with the wind, in miles per hour, is 400 + 50 = 450. 

Since distance = rate • time, or d = rt, we have - = t. Hence,, 

x 
;jrrr = time, in hours, required to fly from A to B, 

and 

x 
jrjr = time, in hours, required to fly from B to A. 

Therefore, from the statement of the problem, it follows that 

x x __ 1 

350 "450 "3* 

Solving this equation, we have 

9x - 7x = 1050, or x = 525. 

So the distance from A to B is 525 miles. 



EXERCISE 1-10 

Solve for the unknown in each problem from 1 to 15. 
1. - 4x - 2 - 3 - 2x. 2. 3x - 6 = x + 12. 3. 3x - 2 = - 4z - 5. 

4. Sy + 7 = 2 - 2y. 5. - 9x - 7 = 6 - — « 

7. 4 - 3z = 6(1 + 2s). 8. 4x - 6 = 3x + 4. 

10. 62/ + 7 = 52/ + 6. 11. 10z - 3 = 9x + 4. 

13. 5z - 7 - Sx = 4z - 17 - 6z 
14. lty + 4 + 6y = 2y + 7 + 3y. 15 

Solve for y in terms of x in each problem from 16 to 24. 
16. 2x - y = 3. 17. 23z + 2y = 4. 18. a: - 4?/ + 10 = 0. 

19. 4y - 2z + 2 = 0. 20. - + f = 1. 21. 6x - 2y = 3. 

0,0 A 

22. 3(x - 4) + 4y = - 3. 23. 2a? + 4(y - 3) = 5. 24. 3x + 7(y - i) = | . 

Solve for all values of x which satisfy the equation in each problem from 25 to 48. 
25. 5x - 3 = 4(s - 2). 26. 8te - 2^ - 9(* - 4) = 13. 27. ^±1 = n . 
°~ ^ 17 9 - 4z _ 4 3Q 5 +6a; _ _ 



5 
6. 4w + - 


= 3»-|. 


9. a? - 


8 = 


: 2s + 3. 


12. \x - 

4 


5 
2 : 


3 1 

= — x • 

2 4 


2?. 

5 + 4x) 


- 3(* - 4) = 0. 



28. 3 _ 7. 

„ a; +5 _s +7_ 
4 5 


29. 
32. 


OJ 2a: - 1 a; — 5 

34 - 3 = 7 • 


35. 


„ 8* - 21 , 1 _ 5x 
5 4 6 




»s-i = e(-J). 





6 3 ""' 3 

x - 6 _ s - 8 »„ x +3 _ 2a; - 7 

"T"-^^' 33 -~5~-~T2~' 
2a; +3 3a: - 1 7a; +3 2 , 1 

~4 2~ =3 - 36 --6- = 5 X+ 2' 

7 - x 4x + 3 6* 16 



38. 



4 7 5 3 



40 -^2x +3 = 7 - 



48 Introductory Topics Sec. 1-30 

* + x 1 x x' ' x 9 9 x ' 

,„ 4 5 1 ., 1 . 3 4 

2x + 3 + x - 4 4x* - lOx - 24 
3 1 

45 * 6x2 _ 2x + 1 ~ 2^17+7 = °- 46, 

47. ^ 7 ~* 



x z - 1 """# + 1 




a? - 1 19 - 22# 


a: + 1 


x +2 z 2 - z - 6 


x - 3 


5 - 2x - .t 2 





6(x - 3) T 2(x 2 + 3x + 9) 81 - 3x 3 
48> 4 + 4+4»_ -5 



x x 2 + 4:X x + 4 

49. Divide 98 into two parts such that one of them exceeds the other by 18. 

50. Find three consecutive integers whose sum is 84. 

51. Find two consecutive integers whose squares differ by 13. 

52. If 8 times a certain number is 9 more than 5 times the same number, what is 
the number? 

53. A rectangular plot of ground is four times as long as it is wide. If its perimeter 
is 4,800 feet, what is its area? 

54. How many pounds of coffee at 90 cents per pound and how many pounds at 
98 cents per pound will it take to make 100 pounds of a mixture costing 96 
cents per pound? 

55. At a college play, admission was 25 cents for a child and 75 cents for an adult. 
If $210 was taken in from 500 admissions, how many children and how many 
adults were admitted? 

56. A man has $365 in 41 bills of $5 and $10 denominations. How many bills of 
each denomination does he have? 

57. What are the angles of a triangle, if one angle is three times the second angle 
and six times the third angle? 

58. One man, X, can do a certain job in 7 days, and another man, Y, can do the 
same job in 15 days. How long would it take them to do the job working 
together? 



2 



The Function Concept 



2-1. RECTANGULAR COORDINATE SYSTEMS IN A PLANE 

In Section 1-5 we saw how we can associate a real number with 
every point on a number scale. The real number attached to a given 
point is called the coordinate of the point. This representation sug- 
gests the assumption that to any real number there corresponds 
precisely one point on the scale, and to any point of the scale there 
corresponds precisely one real number. This one-to-one correspond- 
ence between the set of real numbers and points on the number 
scale is known as a one-dimensional coordinate system. 

We shall now extend the concept of a one-dimensional coordinate 
system to a system of coordinates in a plane in which two number 
scales are perpendicular to each other. The two perpendicular 
lines, which we shall call coordinate axes, divide the plane into four 
parts, or quadrants, numbered as shown in Fig. 2-1. The horizontal 
and vertical lines are designated as the 
x-axis and the y-axis, respectively, and 
their point of intersection is called the 
origin and is labeled O. H 

On each of these axes, we construct a 

number scale by selecting an arbitrary ^ ►JT 

unit of length and the origin as the zero 
point. As in Section 1-5, a coordinate on jjj jy 

the #-axis will be considered positive if 
it is to the right of 0, that is, to the 
right of the y-axis, and will be negative f ig# 2-1. 

if it is to the left. A coordinate on the 

y-axis will be considered positive if it is above the #-axis, and nega- 
tive if it is below. 

Just as the real-number scale of Fig. 1-1 gave us a system of 
one-dimensional coordinates by which we could set up a one-to-one 
correspondence between points on a line and real numbers, so the 

49 



Q 

ii ♦ i 



50 The Function Concept Sec. 2-1 

system of coordinates with respect to two mutually perpendicular 
axes sets up a one-to-one correspondence between points in a plane 
and ordered number pairs. We use the designation "ordered" pairs 
for the following reason. To designate any point, we shall agree to 
give its directed distance frdm the i/-axis first, and call it the 
abscissa or x-coordinate, and then the directed distance from the 
#-axis and call it the ordinate or y -coordinate. The abscissa and 
ordinate of a point constitute its rectangular coordinates. They are 
written in parentheses as an ordered number pair, as in the nota- 
tion (x, y)> the abscissa always being written first. By this scheme 
we assign to each point of the plane a definite ordered pair (x, y) 
of real numbers and, conversely, to each ordered pair (x, y) of 
real numbers there is assigned a definite point of the plane. 

Thus, the abscissa of the point P in 
y Fig. 2-2 is 3, and its ordinate is 2, and 

we say that the coordinates of the point 
P(3,2) P are (3, 2). Similarly, the coordinates 

T of Q are (—2, 0), the coordinates of R 

ni l — ►* are (0, —3), and those of the origin are 
(0, 0). Marking in the plane the posi- 
. hR tion of a point designated by its coordi- 

nates is called plotting the point. 

The coordinate system we have con- 
Fig. 2-2. stfucted is a particular case of cartesian 

coordinates, so called in honor of Rene 
Descartes (1596-1650), who first introduced a coordinate system 
in 1637. It is called a rectangular system, since the axes intersect 
in a right angle. (Actually the axes may intersect at any angle, but 
it is usually simpler to take them perpendicular to each other. When 
the two axes are not perpendicular, the coordinate system is called 
an oblique coordinate system. Oblique systems will not be used in 
this book.) 

2-2. DISTANCE BETWEEN TWO POINTS 

It was seen in Section 1-10 that |a — b\ equals the distance 
between two points on the number scale represented by the real 
numbers a and 6. It follows that \x 2 — Xi\ represents the distance 
between the points A (x u 0) and B(x 2 , 6) on the #-axis of Fig. 2-3. 

Let us now consider the two points P\ and P 2 with coordinates 
(x lf y x ) and (x 2 , yi). The points have the same ^/-coordinate, which 
means that they lie on the same horizontal line. Hence, the dis- 
tance between the points Px(x u Vi) and P 2 (x 2 ,yi) is the same as 



Sec. 2-2 



The Function Concept 





iK 


*2l*vY\) 


* i 
« 

in i 


— i 

i 
i 

1 w V 


A(x v 0) 




B(x 2 ,0) ~ 




Fig. 2-3. 



Fig. 2-4. 



the distance between A(x lf 0) and Z?(# 2 ,0). This distance is 
\x 2 - x±\. Similarly, we can show that \y 2 - Vi\ represents the dis- 
tance between two points (x 2 , Vi) and (x 2 , y 2 ) on a vertical line. 
We have thus arrived at the following two important properties. 
If two points Pi(x u y x ) and P 2 (x 2 , yi) have the same 2/-coordi- 
nate, then the distance between them, or |PiP 2 |, is given by 
(2-1) |PiP 2 | = |x 3 — xi|. 

If two points Qi(x lt y x ) and Q 2 (x u y 2 ) have the same ^-coordi- 
nate then the distance between them is given by 

(2-2) IQ1Q2I = I2/2 ~ 1/1 1 . 

The concept of the distance between any two points in a plane is 
so important that we shall now develop a formula for it. Let us 
denote by d the distance between the points Pi(x u Vi) and 
P 2 {x 2y 2/2). That is, d is the length of the line segment PiP 2 in Fig. 
2-4. Let P 3 be the point (x 2 , y x ) as shown. 

Since the angle at P 3 is a right angle, we have, by the Pytha- 
gorean theorem, 

IP1P2I 2 = IP1P3I 2 + IP3P2I 2 . 
Therefore, 

|PlP2| 2 = |*2 - Xi\ 2 + |tf2 ~ »l| 2 

= (a?2 ~ *i) 2 + (2/2 ~ 2/1) 2 . 

That is, the distance d between any two points Pi (#1,2/1) and 
^2(^2, 2/2) in the plane is given by 

(2-3) <Z = V>2 - *i) 2 + (2/2 - 2/1) 2 . 

This formula is known as the distance formula. 



Example 2-1. Find the distance between the points (— 3» 2) and (5, 2). 

Solution: It makes no difference which point is labeled Pi. Let us label the first 
one Pi and the second one P 2 . Since the two points have the same ^-coordinate, 
(2-1) applies and \PiP 2 1 = |5 - ( - 3)| = |5 + 3| = |8| = 8. 



52 The Function Concept Sec. 2-2 

Example 2-2. Find the distance between the points (3, 1) and (3, 7). 

Solution: Again let us label the first point Pi and the second one P 2 . Since the 
points have the same ^-coordinate, (2-2) applies and IP1P2I = |7 — 1| = 6. 

Example 2-3. Find the distance between the points (2, - 1) and (5, 3). 

Solution: Let us designate the points as Pi(2, - 1) and /Mo, 3). By (2-3), 
the distance IP1P2I is 

d = y/(b -2)2 +(3 -(-l)) 2 



= V9 + 16 = V25 = 5. 

We shall now consider a special application of the distance 
formula (2-3). The line segment OP from the origin to a point 
P(x,y) is called the radius vector to P. By (2-3), the distance 
between O(0, 0) and any point P(x, y), or the length of the radius 
vector to P, is 



or d = V(x - 0) 2 + (y - 0) 2 

(2-4) d = V* 2 + 2/ 2 - 

Thus, for the point P(3, 2), the radius vector has the length 

EXERCISE 2-1 

1. Plot the following points: 

a. (3,5). b. (-4,7). c. (5, -2). 

d. ( - 3, - 6). e. (1, - 3). f. ( - 8, - 6) 
g. (0,2). h. (-5,0). i. (3,0). 

2. In each of the following cases, plot the pair of points and find the distance 
between them: 

a. (2, 3) and (7, 3). b. (5, - 2) and ( - 1, - 2). 

c. (1, 4) and (1, 0). d. ( - 3, - 2) and ( - 3, 4). 

e. (2, 1) and (5, 6). f. (0, - 8) and (5, - 3). 

g. (3,2) and (5, 7). h. ( - 4, - 6) and ( - 8, - 6). 

3. Find the length of the radius vector to each of the following points : 

a. (4, 3). b. (12, 5). c. (1, - 1). 

d. (5, -12). e. (7,0). f. (-3, -2). 
g. (0, 4)._ h. (-1,2). i. (a, 6). 

j. (W3). k. (-\/3, -1). 1. (ro,n). 

2-3. FUNCTIONS 

So far we have been concerned with single numbers and pairs of 
numbers. Now we shall consider mathematical relations, known 



Sec. 2-3 The Function Concept 53 

as functions, between two sets of numbers. To distinguish between 
the two sets, we shall call one of these sets the domain of definition 
X of the function, and the other the range set Y of the function. 

We begin by defining a variable to be a symbol which may take 
any value in a given set of numbers. If # is a symbol which is used 
to denote any number of the domain X and y is a symbol which 
denotes any number of the set Y, then x is called an independent 
variable and y is called a dependent variable. If a set contains only a 
single number, the symbol used to represent that number is called 
a constant 

To set forth a function, the domain should be explicitly specified ; 
that is, it is necessary to determine definitely just what elements 
or numbers the domain contains. The same is true of the range set 
Then, as soon as a definite rule of correspondence is given which 
assigns to each number x of the domain one or more numbers y of 
the range set, the function is specified completely. We thus have a 
set of ordered pairs of numbers (x, y), where x is any number of X 
and y is a number of Y. 

The set of ordered pairs of numbers (x 9 y) is called the function. 
The rule of correspondence which determines the collection of pairs 
(x,y) is often expressed by a formula involving algebraic or other 
processes. In such cases we usually find it convenient to refer to the 
formula as though it were the function. For example, we often 
speak of the function y = x 2 — Sx + 5 when actually we mean the 
set of ordered pairs (x,y) determined by y = x 2 — Sx 4- 5. If just 
one number of Y is paired with each value of x, the function is said 
to be single-valued. If more than one number of Y is paired with 
some value of x, the function is said to be multiple-valued. We fre- 
quently find it possible to deal with multiple-valued functions by 
separating them into distinct single-valued functions. 

The range of values of the function consists of those numbers y 
in the range set Y which actually correspond to some number x of 
the domain. When the range of a function has been determined, 
it is always possible to replace the original range set, which may 
include numbers in addition to those of the range, by the range 
itself. We shall now consider several examples of functions. 

Illustration 1. The constant function y = c associates the same 
number c with every number x of the domain X. Hence, the range 
Y of the function consists of just one number c. Since y has the 
same value for all pairs (x, y), the function is evidently single- 
valued. 

Illustration 2. The identity function y — x associates with every 
real number x the number itself. In other words, the numbers 



54 The Function Concept Sec. 2-3 

corresponding to x = 1, 2, 3, • • • are y = 1, 2, 3, • • respectively. The 
domain X is the set of all real numbers, and the range Y is also 
this entire set. The function y = x is single-valued, because it asso- 
ciates just one number of Y with each value of x. 

Illustration 3. Let us consider the linear function defined by the 
equation y = 3x — 2. The domain X is the set of all real numbers, 
and the range Y is also this entire set. In this case a given x deter- 
mines a unique y which is equal to Sx - 2. For example, corre- 
sponding to x = 1, 2, 3, we have y = 1, 4, 7. The function is single- 
valued. 

Illustration 4. Let the rule of correspondence be given by the 
equation y = x 2 . Also, let X be the set of all real numbers, and let Y 
be the set of all non-negative real numbers. Then, to the number 
x = -2 there corresponds the number y = (-2) 2 - 4; to the number 
x = 3 there corresponds the number y = 9 ; and so on. Hence, y = x 2 
associates just one number of Y with a given value of x, and defines 
y as a single-valued function of x. Although this is not obvious, the 
range of the function is the given range set Y. 

Illustration 5. In this case let the rule of correspondence be given 
by the equation y 2 = x. Here X is the set of all non-negative real 
numbers, and Y is the set of all real numbers. Then to the number 
x = 2 there correspond the two numbers y = \/2 and y = — \/2 ; to 
the number x = 9 there correspond the numbers y = V9 = 3 and 
y = — \/9 = —3 ; and so on. Thus, y 2 = x defines t/asa two-valued 
function of x. Only for x = is there a single value of y, namely, 
y = 0. 

Illustration 6, The function y = y/a 2 — x 2 , with domain X con- 
sisting of all real numbers x such that — a ^ x ^ a, is a single- 
valued function with range set Y consisting of the numbers 
^ y ^ a. Here the range is Y, as may be proved. 

Note that y/a 2 - x 2 has no meaning when a 2 — x 2 is negative. 
Hence, those values of x must be excluded for which a 2 — x 2 < 0, 
and we must restrict the value of x to the interval — a ^ x ^ a. 
(We shall consider the meaning of the square root of a negative 
number in Chapter 11.) The function y = y/a 2 — x 2 is single-valued 
because y/a 2 — x 2 represents the non-negative square root of a 2 — x 2 . 
(The other square root of a 2 — x 2 is denoted by —y/a 2 - x 2 .) 

Illustration 7. The function y = 1/x is defined for all real num- 
bers different from 0. For x = 0, 1/x is not defined, since division 
by zero is not permissible. In other words, although there are 



Sec. 2-4 The Function Contept 55 

values of y for values of x neat 1 0, there is no possible value for y 
when x actually equals 0. 

Usually the functions which we are about to consider are defined 
for all values of x, with the following two exceptions : 

Values of x must be excluded which involve even roots of 
negative numbers, since these are not defined as real numbers. 

Values of x must be excluded for which a denominator is zero, 
since division by zero is not a permissible operation. 

On occasion, the use to be made of a function will restrict the 
values of x for which it is to be regarded as defined. 

2-4. FUNCTIONAL NOTATION 

Since functions are mathematical entities, they may be given 
letter notations, such as /, g, <J>. To designate the number, or num- 
bers, y corresponding to a given number x according to the rule 
specified by a given function /, we us6 the notation /(#). As an 
illustration, let the function / be defined by the equation 

y =* x 2 — 2x + 3. 

Then /(0) = 3, /(-l) = 6, and so on. frequently, the symbol f(x) 
is used to designate the function rather than the functional values. 
The context will make the meaning cleai*. 

It should be remembered that the notation y = f(x) does not 
mean that y is a number / multiplied by another number x. Instead 
it is an abbreviation for "f of x." 

The set X does not have to be as simple as in the preceding illus- 
trations. If X should consist of a set of ordered pairs of numbers, the 
rule of correspondence would then determine a value, or values, of y 
for each ordered pair of X. We would then have a function of two 
independent variables. For example, the area of a triangle is given 
by the relationship 

A = /(a, b) = gab. 

Here X is the set of ordered pairs, (a, b), of positive real numbers, 
where a is the length of the altitude of the triangle and 6 is the 
length of its base ; and Y is the set of numbers A, each of which 
represents an area corresponding to a given pair (a, 6). Similarly, 
a function of three variables, f(x, y, z), is defined in terms of a set 
X of ordered triples (x,y,z). Thus, if f(x, y) = x 2 + y 2 , then 
/(2, 3) = 2 2 + 3 2 = 13. Also, if f(x, y> z) ~ x - y + 2z, then /(3, 2, 5) 
= 3-2 + 2(5) =11. 



56 The Function Concept Sec. 2-4 

EXERCISE 2-2 

By using the phrase "a function of" in each problem from 1 to 8, express each 
given quantity, which is regarded as a dependent variable, as a function of one or 
more independent variables. Where possible write the relationship both in words 
and in symbols. 

I. The area of a circle. 2. The area of a triangle. 
3. The area of a trapezoid. 4. The volume of a sphere. 

5. The volume of a cylinder. 6. The retail price of food in a grocery store. 

7. The annual premium for a life insurance policy. 

8. A person's height. 

9. Given /(as) = 2x - 3, find /(0), /(- 1), /(3), /(1/2), /(V§), 3/(1), /(3)/4, 
1 

'fix)' 

10. Given g(x) = 3x 2 + 5, find g(l), g(- 3), (p(2))«, g(4), 2?(4), <;(* - 1). 

/( — 2) 

II. Using f(x) and g(x) as defined in problems 9 and 10, find — y^r- > /(6)gf(3), 

In problems 12 to 26, let /(#, 2/) = 2xy + 3.r - 2?/, and let g(a> 6, c) 
= a 2 -f 6 2 + c 2 . Evaluate or simplify each given expression. 

12. /(I, 2). 13. /(O, 0). 14. gf(0, 0, 0). 

15. </(l, 2, 3). 16. /(*, 1A). 17. 0(p, <?, r). 

18. fix, y) + gix, y, ,). 19. A°L . 20 . /^g . 

/(if,*) . 



/(</),/© 



21. /(I, - 1) • gW% V3, V5). 22. 



?(& x, z) 



2 %(3,2,1) rf-3,-2,-1) **' 9( a > b > C) - 

25. #(- a, b, c) - #(a, - b, c). 26. #(a, 6, - c). 

27. Express the area A and circumference of C of a circle as functions of the radius r. 
By eliminating r, express A as a function of C, and also express C as a func- 
tion of A. 

28. Express the volume of a sphere as a function of its surface area. 

29. Express the surface arfea of a sphere as a function of its volume. 

30. Suppose that U is a function of V and that V is a function of W. Show that U 
is a function of W. 

Determine the maximal domain of values of x for which y is defined as a function 
of x in each problem from 31 to 61. Assume that x and y are real numbers. 

31. y = a. 32. i/ = 3z. 33. 2/ = - | • 
34. y = 3z + 1. 35. 1/ = 2a; - 3. 36. 2/ = 4x + 5. 
37. 2/ = z 2 . 38. 2/ = z 2 - 2. 39. 2/ = z 2 + 1. 

40. y = *(2:r + 1). 41. y = (3s - 1) (a? + 1). 42. ?y = (2x - 3) (2a? + 1). 

43. y = x(s - 1) (x + 1). 44. 2/ = z 2 + - • 45. y = 



g * s(x - 1) 

46. x 2 + 2/ 2 = 9. 47. x 2 - y 2 = 9. 48. z 2 - y 2 = 0. 

z 2 



49. x 2 + y 2 = 0. 50. 3s 2 + 2/ 2 = 0. 51. y = 



1 + z 2 



Sec. 2—6 The Function Concept 57 

1 11 x 2 

52. y = ^ • 53. y = — — - - - • 54. y = — r-r • 

* 1 - s * x + 2 x * x 4- 1 

55 ' 2/ = *! 7 I * 56. x = 4i/ 2 . 57. x + 1 = 32/ 2 . 

58. !/ = Vl - z 2 . 59. y = \/4 - a; 2 . 60. y = \/z 2 + 1. 

2-5. SOME SPECIAL FUNCTIONS 

The following additional illustrations of two rather unusual, but 
very useful, functions are given here to help us become better 
acquainted with the function concept. 

The absolute-value function is defined by associating with x its 
absolute value \x\. The functional relation is given by y—\x\. 
Thus, the domain comprises the set of all real numbers, while the 
range comprises the set of all non-negative real numbers. For this 
function we have the values 2, 3, 0, ir, y/2 corresponding, respec- 
tively, to x = 2, —3, 0, 77, — \/2. 

The bracket function or greatest-integer function, represented 
by the notation y = [x] , is defined as the largest integer which does 
not exceed x. Its domain is the set of all real numbers, and its 
range is the set of all integers. Thus, if f(x) = [x], then /(— 3.5) 
- -4, /(-l) = -1, /(0) = 0, /(2.5) = 2, and /(5) = 5. 



EXERCISE 2-3 

1. Given /(x) = [x | , find/(- 3),/(2.3), and /(/(*))• 

2. Given /(x) = [x], find/(3.2),/(2), and/(- 3). 

3. Given /(x) = x r M> find/(- 3) and/(2.5). 

4. Given /(x) = | x | + [x] - x, find/( - 3.5) and/(4). 

5. Given /(x) = [x] + [2 - x] - 1, find/(0),/(- l/2),/(l),/(3/2), and/(2). 

6. Let/(x) be the function whose domain X is the set of all real numbers for which 
the definition is as follows : 

if x < 0, then/(x) = - x; 
if x ^ 0, then fix) = x. 
Find/(3)and/(-2). 

2-6. VARIATION 

A particularly important example of a simple type of function 
often occurring in the physical sciences is given by the formula 

y = kx. 

If k > 0, this equation shows that y increases a§ x increases, and 
that y decreases as x decreases. We usually say that y varies directly 



58 The Function Concept Sec. 2—6 

as x, or that y is directly proportional to x. If x ¥= 0, the relation- 
ship may also be written as follows : 

x . ' 

where k is called the constant of proportionality. This relationship 
is equivalent to saying that the ratio of y to x remains constant for 
all non-zero values of x. 

The value of the constant k in any particular problem may be 
determined from a known pair of values of x and y in the problem. 
Thus, if the given relationship is y = kx, and if we know that y = 6 
when x = 2, then k = 3. The formula then becomes y = Sx. 

We say that y varies inversely as x, or y is inversely proportional 

to a, if h 

y=l (x * 0). 

This relationship shows that y decreases as x increases, and that y 
increases as x decreases. But, when x ¥= 0, the following two for- 
mulas are equivalent : ^ 

xy = k and y = - • 

Therefore, the relationship between a? and 2/ is such that the product 
of # and y is constant. 

Several types of variations may be combined in a single equation 
to express a certain law. For example, when y varies directly as x 
and z, we say that y varies jointly with x and z, and we write 

y = te. 

Direct variation and inverse variation are often combined in 

applications. Thus, according to Newton's law of gravitation, the 

force F of attraction between two bodies of masses mi and m 2 varies 

directly as the product of their masses and inversely as the square 

of the distance d between th$m. The equation is 

„ _ kmim2 
F--35-. 

Example 2-4. If a man is paid $15 for an 8-hour day, how much would he make 
in a 35-hour week? 

Solution: The wages a man earns vary directly as the amount of time he works. 
Since this is a problem in direct variation, we have 

w = kt. 
In this formula, w represents the total wages, in dollars; t represents the time 
worked, in hours; and the constant k represents the wage rate in dollars per hour. 
Substituting w = 15 and t = 8 determine^ the constant k = 1.875. The general 
formula then becomes 

w = 1.875 t. 



Sec. 2-6 The Function Concept 59 

Therefore, when t = 35, we have 

w = (1.875) (35) = 65.62. 
Hence, the man earns $65.62 in a week. 

Example 2-5. A motorist traveling at an average rate of 50 miles per hour made 
a trip in 5 hours. How long would it take him to make the same trip at an average 
rate of 60 miles per hour? 

Solution: Since the time required Varies inversely as the speed, we have 

r 
In this case, k represents the distance traveled, in miles; r is the speed, in miles 
per hour; and t is the time required, in hours. Substituting t = 5 and r = 50 
determines k = 250. Hence, the general formula is 

t = — • 
r 

Therefore, when r = 60, we obtain 

250 _ 25 

60 6 * 
Hence, the time required is 4 hours 10 minutes. 

Example 2-6. Under suitable conditions the electric current / in a conductor 
varies directly as the electromotive force E and inversely as the resistance R of the 
conductor. When E = 110 volts and R = 10 ohms, / = 11 amperes. Find what 
voltage is necessary to cause 2 amperes to flow through 60 ohms of resistance. 

Solution: From the statement of the problem, we see that the combined variation 
is given by the formula ^ 

1 "If 

Substituting I = 11, E — 110, and R = 10 determines the constant k to be 1. 
Therefore, the formula becomes 

R 

This relationship may also be expressed as follows: The required voltage E 
is equal to the current / flowing through the conductor multiplied by the resistance 
B, or E = IR. 

For the specified values, E = (2) (60) = 120. Hence, E = 120 volts. 

In this problem k = 1. The formula obtained is commonly known as Ohm's law. 
It is widely applied to entire and partial circuits through which electric currents flow. 



EXERCISE 2-4 

1. If y varies directly with #, and y = 15 wh&i x = 7, find a formula for y in 
terms of x, 

2. If x varies directly with y, and x *= 32 when y = 4, find x When y = 3. 

3. If y is directly proportional to x 2 > and y = 112 when a; = 4, find y when a; = 9. 

4. If y is inversely proportional to x, and 2/ = - % when x = 1, find y when £ = -3. 



60 The Function Concept Sec. 2-6 

5. If y varies inversely with x ) and y = 10 when x = 3, find y when a; = 6. 

6. If x varies directly with y and inversely with 2, and a? = 4 when y = 12 and 
2=2, find # when ?/ = 16 and z = 4. 

7. If y varies directly with y/x and inversely with z 2 , and 2/ = 18 when x = 9 
and 2=2, find 2/ when £ = 25 and 2 = 6. 

8. If t/ is directly proportional to x and inversely proportional to y/z } and 2/ = 4 
when # = 1 and 2 = 1, find ?/ when x = 2 and 2=4. 

9. If ?/ varies directly with x z and inversely with 1 — 2 2 , and y = 2 when a; = 1 
and 2=2, find 2/ when x = — 1 and 2 = — 2. 

10. If two spheres have radii n and r 2 , diameters di and ^2, and surface areas Si 
and £ 2 , respectively, show that 

r-2 2 ~ d 2 2 ~~ S 2 * 

11. If the two spheres in problem 10 have volumes V\ and F2, respectively, show 
that 

Vx ^rj* di» 
7 2 r 2 3 d 2 3 ' 

12. If the radii of two spheres are 3 units and 1 unit, respectively, find a) the ratio 
of their surface areas and b) the ratio of their volumes. 

13. What are the ratio of the surface areas and the ratio of the volumes of two 
spheres if the ratio of their radii is 3/2? 

14. The diameter of the planet Jupiter is approximately 10.9 times the diameter of 
Earth. Assuming that both planets are spheres, find a) the ratio of their surface 
areas and b) the ratio of their volumes. 

15. The diameter of the sun is approximately 109 times the diameter of Earth. 
Compare the volumes and the surface areas of the sun and Jupiter, if we assume 
that the sun and both planets arc spheres. 

16. By how much must the diameter of a sphere be multiplied to give a sphere 
whose surface area is 25 times that of a given sphere? 

17. When the volume V of a gas remains constant, the pressure P varies directly as 
its absolute temperature T. (Absolute temperature is measured from the 
so-called absolute zero, which is approximately — 460° F or — 273° C.) If gas 
is enclosed in a tank having a volume of 1,000 cubic feet and the pressure is_ 
54 pounds per square inch at a temperature of 27° C, what will be the tempera- 
ture when the pressure is raised to 108 pounds per square inch? 

18. If the temperature T of a gas remains constant, the pressure P varies inversely 
as the volume V. A gas at a pressure of 50 pounds per square inch has a volume 
of 1,000 cubic feet. If the pressure is increased to 150 pounds per square inch 
while the temperature remains constant, what is the volume? 

19* The weight of a body above the earth's surface varies inversely as the Square 

of the distance from the center of the earth. If a certain body weighs 100 

pounds when it is 4,000 miles from the center of the earth, hdw much will it 

weigh when it is 4,010 miles from the center arid when 4,100 miles from the 

• center? 



Sec. 2-7 The Function Concept 61 

20. The electrical resistance of a wire varies directly as its length and inversely as 
the square of its diameter. A copper wire 10 inches long and 0.04 inches in 
diameter has a resistance of 0.0656 ohms, approximately. What is the resistance 
of a copper wire 1 inch long and 0.01 inches in diameter? 

21. What is the diameter of a copper wire 1,000 inches long whose resistance 
is 10 ohms? 

22. According to Kepler's third law, the square of the time it takes a planet to 
make one circuit about the sun varies as the cube of its mean distance from the 
sun. The mean distance of the earth is 92.9 million miles, and the mean 
distance of Jupiter is 475.5 million miles. Find the time it takes Jupiter to 
make one circuit about the sun. 

2-7. CLASSIFICATION OF FUNCTIONS 

It is often desirable to group functions into classes. For our 
immediate purpose it will suffice to consider a classification into 
algebraic functions and non-algebraic, or transcendental, functions. 
Let us first give a more precise definition of a polynomial function 
and then define algebraic functions and give some illustrations 
of both. 

A polynomial function of x is a function given by the relationship 

y = a x n + aix n ~ l + h a n _ix + a n , 

where a , a l9 • • • , a n - l9 a n are real constants, a ¥= 0, and wis a posi- 
tive integer or zero. The polynomial function is said to be of 
degree n. The function which makes the number correspond to 
every number x is also called the zero polynomial, but this poly- 
nomial has no degree. 

A rational function of x is a function which either is a poly- 
nomial function or can be expressed as a quotient of two poly- 
nomials. Thus, a polynomial is often referred to as a rational 
integral function of x. 

A polynomial, or a rational integral function, of x, y, z, ■ • • , is 
defined to be the algebraic sum of terms of the form 

kx a y b z c ■ • • , 

where & is a constant coefficient and each of the exponents a, b, c, 
•is either a positive integer or zero*. The degree of such a func- 
tion is the degree of the highest-degree term which is present. 
For example, the expressions 3x 2 - 5 and 5x 2 — 7xy 2 + &z define 

polynomial functions of the second degree and third degree, respec- 

<■»• — in . — oj 

tively. These and the expressions —~ and 2x — v 7 + 2 T i are 

examples of rational functions. 

* Zero exponents will be defined in Chapter 4. For now, one needs only note 
that u° = 1 for any non-zero number u. 



62 The Function Concept Sec. 2-7 

A number is an algebraic number if it is a root of a polynomial 
equation of the form 

a x n + aix n ~ l + • • • + a n _i# + a„ = 0, 

in which the coefficients Oo, &i, ■ • • a n are integers, not all zero. 

Analogous to the term algebraic number, we have the term 
algebraic function. A function y = f(x) is called an algebraic func- 
tion of x if y is a solution of an equation of the form 

Po(x)y n + ^lWr 1 + • • • + P n -i(x)y + P n (x) = 0, 

where the coefficients P (#), Pi(#), ■ * • , P*(#) are polynomials in 
x, and n is a positive integer. 

Polynomials and rational functions are special types of algebraic 
functions. The functions that we have considered so far were illus- 
trations of algebraic functions. According to our definition, 
y = V# is an algebraic function of x because y 2 — x — 0. In this 
case, n = 2, P (x)=l, P 1 (x)=0, and P 2 (%) = x. Similarly, 

is an algebraic function of x because xy 2 - x 2 + 1 = 0. 



y=^ x - 



X 

Here n = 2, P (:r) = a?, P^a) = 0, and P 2 (z) = - x 2 + 1. 

An irrational function is an algebraic function which is not a 
rational function. 

A transcendental number is a number which is not algebraic, 
and a transcendental function is a function which is not algebraic. 

Functions like the trigonometric functions, which we shall take 
up in Section 3-2, belong to the class of transcendental functions. 
Later we shall consider other types of transcendental functions, 
namely, the logarithmic and exponential functions such as log x 
and 10*. 



O The Trigonometric Functions 



3-1. THE POINT FUNCTION P(») 

The trigonometric functions that we are about to define are func- 
tions in the sense previously described in Section 2-3 ; that is, they 
are relations between two sets of numbers. The student who is 
familiar with the trigonometric functions from his high-school work 
is cautioned to note that we are not, for the present, discussing angles 
in connection with these functions. We shall see that the concept 
of a trigonometric function need not be associated with an angle; 
in fact, many of the most important applications of mathematics 
in modern science and engineering are concerned with trigonomet- 
ric functions of pure numbers. Hence, we shall adopt the numer- 
ical point of view, leaving the study in terms of angles as a 
secondary consideration. 

Consider a circle with a radius of one unit placed at the origin 
of a rectangular-coordinate system. See Fig. 3-1. Let t be any real 




*»x 




Fig. 3-1. Fig. 3-2. 

number. Starting at the point with coordinates (1,0), we lay off 
on the unit circle an arc of length \t\. If t> 0, we measure the arc 
in a counterclockwise direction. If t < 0, we measure in the clock- 
wise direction. If t = 0, the arc consists only of the point (1,0). By 
this procedure, there is associated with each real number t a defi- 
nite end-point P(t) of the arc whose initial point is (1, 0). There- 

63 



64 The Trigonometric Functions Sec. 3—1 

fore, corresponding to every real number t, we have a definite 
ordered pair (x, y) of numbers which are the coordinates of the 
endpoint of the arc. 

Since P(t) lies on the unit circle, it is at one unit distance from 
the origin. Hence, it follows from the distance formula that 
(3-1) x 2 + y 2 = 1. 

By means of this equation we can find the second coordinate of the 
point P(t), except for sign, if one of the coordinates is known. 

To determine the number pair (x, y) for the point P(t) corre- 
sponding to a given value of t, we shall take note of the fact that the 
circumference of the unit circle is 2tt — 6.2832 units (approxi- 
mately) . For example, since arc ABC in Fig. 3-2 is one-half of the 
complete circumference, it is tt units in length, and arc AB is equal 
to it/2. It now becomes apparent that P(0) is the initial point 
(1, 0) ; P(tt) is the point (-1, 0) ; P(tt/2) is the point (0, 1) ; and 
jP(3tt/2) and P(-tt/2) both represent the point (0,-1). 

3-2. DEFINITIONS OF THE TRIGONOMETRIC FUNCTIONS 

The correspondence between the set of real numbers t and the set 
of ordered pairs (x, y) leads to definitions of the six common 
trigonometric functions, namely, the sine, cosine, tangent, cotan- 
gent, secant and cosecant. 

General Relationships. We shall define the cosine of the real 
number t to be x, and the sine of the real number t to be y. Thus, 
we have 

(3-2) x = cos t, 

and 
(3-3) y = sin t. 

The domain of each of these functions (the sine function and the 
cosine function) is the set of all real numbers t. Since, however, 
every point P(t) lies on the unit circle, neither of its coordinates 
(x, y) can exceed 1 in absolute value. Therefore, 
(3-4) |cos*|^l and |sin*|^l. 

In other words, the ranges of cos t and sin t are restricted by the 
requirements —1 ^ cos t ^ 1 and —1 ^ sin t ^ 1, respectively, for all 
values of t It may be shown that the range of each of these func- 
tions is the set of all real numbers u such that -1 ^ u ^ 1. 

The other four functions may be defined in terms of the cosine 
and sine, as follows : 

(3-5) tan t = ^4 (cos * * 0), 

7 cos t 



Sec. 3-2 The Trigonometric Functions 65 

(3-6) cot t = 5°L? ( sin * * <>)> 

' sin £ 

(3-7) sec * = -^— (cos < ^ 0), 

COS v 

(3-8) esc t = -X- (sin * ^ 0). 

7 sin £ 

Since the cosine and sine are defined in terms of the coordinates 
of the point P(t), it is also possible to express the other functions 
in terms of these same coordinates. From the definitions of the 
cosine and the sine given by (3-2) and (3-3) and from the defini- 
tions of the other functions given by (3-5), (3-6), (3-7), and 
(3-8), we have 

(3-9) tan t = ^4 = ~ ' ( x * °)> 

v ' cos t X 

(3-10) C ott = ~~ = ~ J (y*0), 

v ' sm t y 

(3-11) sec t = -^-7 = - ^ (a? ^ 0), 

v y cos £ X ~ 

(3-12) esc * = -J— =- ' (y**0). 

smty 

We note here that cos t, or x, appears in the denominators of both 
tan t and sec t. Hence, tan t and sec t are not defined when t is a 
number for which the x-coordinate of P(t) equals zero. For 
example, since the ^-coordinate of P(tt/2) or P(37r/2) is 0, it fol- 
lows that tan 7r/2, sec 7r/2, tan 3tt/2, and sec 3t7/2 are not defined. 
Similarly, it can be shown that cot 0, esc 0, cot it, and esc u are not 
defined. We conclude, therefore, that the domain of each of the 
functions tan t, cot t, sec t, and esc t is the set of all real numbers 
for which the denominator is not zero. 

It also follows from (3-7) and (3-8) that numerical values of 
sec t or of esc t can never be less than 1. Hence, the ranges of 
these two functions are restricted by the requirements 
(3-13) |sec t\ ^ 1 and |csc t\ ^ 1. 

From (3-5) and (3-6) we obtain an idea of the behavior of the 
tangent and cotangent functions. It may be shown that the range 
of each of these funptions is the set of all real numbers. 

The Trigonometric Functions of 7r/6, 7t/4, and 7r/3. The computa- 
tion of the numerical values of the trigonometric functions in gen- 
eral is beyond the scope of this book. However, we shall use the 
methods of plane geometry to find sin t, cos t, and tan t for £.= 7r/6, 



66 



The Trigonometric Functions 



Sec. 3-2 




*»x 



Fig. 3-3. 



Fig. 3-4. 



Fig. 3-5. 



7r/4, and tt/3, in order to show that for certain values of t the 
trigonometric functions can be found exactly without tables. 

To compute the functions for the real value t = 7r/3, we con- 
struct the unit circle of Fig. 3-3. Arc AB is given to be equal to 
7r/3, which is one-sixth of the complete circumference. Triangle 
OAB is inscribed in the circle, as shown, with side BO extended 
through the origin to C. Our problem now is to find the values of 
x = cos (7r/3) and y = sin (tt/3) as coordinates of the point B. 

Since OA = OB, triangle AOB is isosceles. Hence, 

angle OAB = angle OBA. 
We note also that arc CAB = tt and 



arc CA = arc CAB - arc AB = tt - tt/3 = 2tt/3. 



Therefore, 
Also, 



arc CA = 2 arc AB. 
angle CO A = 2 angle AOB. 



Furthermore, angle COA is an exterior angle of triangle AOB. 
Hence, it equals the sum of the two remote interior angles OAB 
and OBA, or 

angle COA = angle OAB + angle OBA. 
Thus, 

2 angle AOB = angle OAB + angle OBA, 

and triangle AOB is equilateral. 

If we draw BD perpendicular to OA, it will bisect OA. Then 
x = 1/2 ; and from # 2 + # 2 = 1 it follows that y 2 = 3/4 and y = VS/2. 
Hence, cos (tt/3) = 1/2, sin (tt/3) = VS/2, and tan (tt/3) = V& 

For t = 7r/6, place the equilateral triangle AOB of Fig. 3-3 in the 
unit circle as shown in Fig. 3-4, where E is the mid-point of arc 
AB. Then arc EB = 7r/6 and OE is the perpendicular bisector of 
chord AB. 



Sec. 3-2 The Trigonometric Functions 67 

Since DB is one-half of chordAB, y = 1/2. From x 2 + 2/ 2 = 1 it 
follows that £ 2 = 3/4 and a = V3/2. Hence, cos (tt/6) = \/5/2 and 
sin (tt/6) = 1/2. We then have tan (tt/6) = 1/y/S = Vff/3. 

For £ = 7r/4, construct a unit circle as shown in Fig. 3-5 with 
arc AB = tt/4. Since arc AC = tt/2, arc 5C = tt/2 — 7r/4 = ir/4. 
Hence, arc AB = arc Z?C, and 

angle A 05 = angle 50C. 

Draw SD perpendicular to OA. Since the two parallels OC and 
DB are cut by the transversal OB, 

angle BOC = angle OBZ)., 
Hence 

angle AOB = angle OBZ>, 
and 

02) = DB. 

Thus, y = x. Substituting this value of y in x 2 + y 2 = 1, we have 
2# 2 = 1 and # 2 = 1/2. Therefore, a; = cos (tt/4) = \/2/2, and y = 
sin (tt/4) = V2/2. It follows-that tan (tt/4) = 1. 

Other Special Values. In a similar fashion we can compute 

exactly the trigonometric functions of such values of t as — > — > 

7tt 3tt 

— i and — - • Functions of multiples of rr/2 may also be computed 
6 4 

in this fashion, if one considers a straight line as a right triangle 

in which one angle is and, hence, one side has zero length. 

Example 3-1. Calculate the values of the six trigonometric functions of t = ir/2. 

Solution: As explained in Section 3-1, the coordinates of the point P(ir/2) are 
(0, 1). Hence, by (3-2), (3-3), (3-9), (3-10), (3-11), and (3-12), 
cos (t/2) = 0, sec (ir/2) is undefined, 

sin (tt/2) = 1, esc (tt/2) = 1, 

tan (tt/2) is undefined, cot (w/2) = 0. 



EXERCISE 3-1 

1. Determine the coordinates of each of the following points: 

a. P(2tt). b. P(-tt). c. P(5x/2,. 

d. p(~y)' e - p ( 4 *)- f - p (- 7 *)- 

2. In each of the following cases, carefully draw a unit circle and estimate the 
coordinates of P(t). 

a. P(l). b. P(2). c. P(3). 

d. P(-2). e. P(4). f. P(-3). 



68 



The Trigonometric Functions 



See. 3-2 



3. Evaluate each of the following: 



. 3tt 

a, sin -r- • 
4 


- 57T 

b. cos -r- • 
o 


c. tan(--~) 


d. COt -r- • 


. llTT 

e. sm -=— • 

6 


f. sec -r- • 




g. csc (- g)« 


h. sm (- y)- 


/ 5tt\ 
.. cob(- t ) 



4. In each of the following cases assume that ^ t ^ 2ir, and draw a figure showing 
approximately the appropriate arc (or arcs). 

a. sin t = 1/2. b. cot i = — 1. c. tan t = 1. d. csc £ = — 1. 

e. cos 2 = — 1/2 > sin t being positive. f. cot t = — 1, sec J being negative. 

5. Complete the following table, which shows the algebraic signs of the trig- 
onometric functions in the four quadrants. 





Quadrant in which P(t) lies 


Function 












I 


II 


Ill 


IV 


cos t 


+ 


— 


— 


■v 


sin t 


+ 


+ 


- 


-~ 


tan t 


+ 


- 


+ 


— 


cot t 










sec t 










csc t 











6. Use the equation x 2 + y 2 = 1 to find all values of t for which tan t = cot t, 
where £ t g, 2tt. 

7. Use the equation x 2 + y 2 — 1 in each of the following cases to find the other 
trigonometric functions of t. 

a. sin t = 1/2. b. cos t = 3/5. c. sec t = 13/12. 

d. csc t = - 3/2. e. sec t = - 2. f. tan i = 4/3. 

g. cot * = 2. h. tan * = - 6/7. i. sin t = - 3/5. 

8. Prove that each of the following equations is correct. 

a. sin ( - t) = - sin t. b. cos ( - t) = cos t. 

c. sec ( — t) = sec £. '- d. tan ( — t) = — tan t. 

9. For each of the following cases, state the quadrant, or quadrants, in which the 
given condition is satisfied. 

a. The sine and cosine have the same signs. 

b. The tangent and cosine have opposite signs. 



3-3. IDENTITIES 

As an immediate consequence of the definitions of the six trigo- 
nometric functions, we can establish certain relationships among 



Sec. 3—3 The Trigonometric Functions 69 

them which hold for every value of t. Since P(t) lies on the unit 
circle, (3-1) holds ; that is, x 2 + y 2 = 1. But, according to (3-2) and 
(3-3) , x = cos t and y = sin t. Therefore, we have the equation 
(3-14) cos 2 t + sin 2 t = 1. 

This states that "the square of the cosine of t plus the square of 
the sine of t equals unity." Since (3-14) holds for every value of t, 
it is an identity. Note that we use the symbol cos 2 t instead of 
(cos t) 2 . This simplified notation is used for all positive exponents, 
but is never used in the case of a negative exponent. Thus, cos -1 t 
does not mean the same as (cos t)~\ as we shall see in Chapter 8. 

Similarly, we can prove that for each value of t for which the 
functions are defined, 
(3-15) 1 + tan 2 t = sec 2 t 9 

(3-16) 1 + cot 2 t = esc 2 t. 

Proof of (3-15): By definition, tan t = - — - and sec £ = ■ 



cos t cos t 

However, these relationships have no meaning if cos t equals zero, 

that is, if the ^-coordinate vf P(t) equals zero. When cos t^O, 

we may divide both sides of the identity cos 2 1 + sin 2 1 = 1 by cos 2 t 

and obtain 

sin 2 t ___ 1 

„, „ cos 2 t "~ cos 2 t 

Therefore, 

1 + tan 2 t = sec 2 t. 

cos t 1 

Proof of (8-16): By definition, cot t = - — - and esc t = 



sin t sin J 

In this case we assume that sin t ^ 0. When we divide both sides 

of sin 2 1 + cos- £ = 1 by sin- £, we obtain 

cos 2 2 1 



sin 2 t sin 2 £ 
Therefore, 

1 + cot 2 t = esc 2 t. 

We thus have established the three identities which we restate 
here for easier reference : 

(3-14) cos 2 t + sin 2 t = 1, 

(3-15) 1 + tan 2 t = sec 2 t y 

(3-16) 1 + cot 2 t = esc 2 L 

These fundamental! identities are very important in working with 
trigonometric identities and should be remembered. 

Our present work with identities will consist of reducing given 
trigonometric expressions to other forms. Unfortunately, no specific 
rule of procedure can be given for making these reductions. Profit- 



70 The Trigonometric Functions Sec. 3-3 

ciency in making such reductions is a matter of both practice and 
experience. Generally speaking, when we want to reduce a given 
expression to some other form, it is helpful first to perform any 
indicated algebraic operations and then to use some form of one of 
the fundamental relationships to simplify the expression. 

To prove an identity, we may proceed in any one of the follow- 
ing ways : 

1) We may work on the more complicated member of the identity 
and attempt to reduce it to the simpler member. 

2) We may work with both sides of the identity and show that they 
induce to the same expression. 

3) We may form the difference of the two sides of the identity and 
prove that difference to be equal to zero. 

It is frequently desirable to express both sides of the given identity 
in terms of sines and cosines alone, and then use (3-14) if needed. 
We shall consider a few examples to illustrate the procedure in 
reducing expressions. 

Example 3-2. Show that cos t + sin t tan t = sec t. 

Solution: From (3-5), or the definition of the tangent, we have 

* , • * * * 4 . • * sin t 
cos t -f sin t tan t = cos t -f sin t • 

Adding, we have . . 9 . , . ., . 

& * M , . M sin t cos 2 t + sin 2 1 

cos t + sin t = • 

cos t cos t 

Since cos 2 1 -f sin 2 1 = 1 and r = sec t, 

cos t 

cos t + sin t tan t = = sec t. 

cos t 

Example 3-3. Reduce — — to sin t cos t. 

r tan t -f cot t 

sin t cos t 

Solution: By definition, tan t = and cot t = - — - • Therefore, 

J cos t sin t 

1 1 1 cos t sin t 



tan t + cot t sin t cos t sin 2 1 + cos 2 1 sin 2 t + cos 2 £ 
cos 2 sin £ cos £ sin t 

Since sin 2 J + cos 2 1 = 1, we obtain 

1 cos / sin t 



tan J + cot t 



= sin £ cos £. 



Example 3-4. Establish the identity (sec J - cos t) 2 = tan 2 J(l - cos 2 by 
reducing both sides to the same expression. 



Solution: By definition, sec 2 = ; • Hence, the left side becomes 

cos t 

Vcos J / \ cos J / 



Sec. 3-4 The Trigonometric Functions 71 

By (3-14) we have x1 9 . 9 , . 9 _ x 9 . . . 

'1 — cos 2 J\ 2 /sin 2 J\ 2 sin 4 £ 



/ l - cos 2 t y__ / sin 2 l \ 2 _ i 
\ cos £ / ~~ Vcos t/ ~~ i 



cos 2 1 

The right side of the given identity may be reduced as follows. By definition, 

. , sin t 

tan t = - • 

cos t 

Hence, we have . n A . M M 

- 9 ,, t 9 A sm 2 1 . sin 4 J 

tan 2 J(l - cos 2 t) = — — • sm 2 1 = — — • 

<?os 2 1 cos 2 1 

Since both sides reduce to the same expression, — — > the identity is established. 

EXERCISE 3-2 

Prove each of the following identities: 

- . , . , , n sin t cos t t 

1. sm t - cos t tan t = 0. 2. 7 H = 1. 

esc 2 sec t 

sinf secf n 4 . . . . 

3. : = 0. 4. tan J esc t = sec £. 

cos t esc £ 

5. sin t(cot t 4- esc = 1 -f cos £. 6. (sin £ — cos 2 = 1 — 2 sin £ cos J. 

7. sec 2 J 4- 2 tan t = (1 -f tan 2 . 8. sin £(csc t - sin = cos 2 1. 

9. tan 2 *(cot 2 1 - cos 2 = cos 2 U- 10. cos 2 * 4- cos 2 1 tan 2 $ = 1. 

11. (1 - sin (1 + sin t) = cos 2 1. 12. (sec t - 1) (sec t + 1) = tan 2 *. 

13. — = esc L 14. sin t tan t + cos 2 £ sec t = sec $. 

1 - cos 2 £ 

15. sin 2 1 esc 2 * = 2 - cos 2 J sec 2 1. 16. sin 4 * sec 2 1 esc 2 J = tan 2 L 

17. tan J + cot t = sec J esc J. 18. esc 2 1 4- sec 2 J = sec 2 1 esc 2 J. 

19. sec 4 1 - sec 2 * = tan 4 1 + tan 2 1. 20. sin 4 J - cos 4 1 = sin 2 J - cos 2 1. 

21. sm * tan * = _ S ec *. 22. tan 2 t + cot 2 « = sec 2 1 csc ? « - 2. 

cos 2 J - 1 

9q 1 — sin ^ 1 — cos < __ sin t + cos I — 1 ^ 

cos £ sin t ~~ sin £ cos £ 

, sec 2 f - tan 2 t 4- tan J , . , , 

24. = sin t 4- cos L 

sec £ 

25. sec 2 1 = esc 2 *(sec 2 1 - 1). 26. (tan t + cot J) 2 = sec 2 1 esc 2 J. 

27. 1 4 tan 2 £ = (sec 2 1 - l)csc 2 «. 28. sin t(\ 4- tan 2 1) - sin * = tan 3 1 cos *. 
sin t — co s I __ tan t — 1 ^ ~ ft sin 3 6 -f cos 3 t 

sin < 4- cos t "~ tan £4-1 sin £ + cos £ 

1 — cos t 
1 + cos t 



^ A sm £ — cos t tan f - 1 ^ A sin 3 1 + cos 3 1 - . . . 

29. -; — . . = . 30. —: — — = 1 — sin t cos I 

sm t 4- c" 

31. cot 2 1 - cos 2 1 = cos 2 * cot 2 1. 32. * 7 C ° 8 ! = ( csc * - cot 2 • 



3-4. TABLES OF TRIGONOMETRIC FUNCTIONS 

Exact numerical values of trigonometric functions in general 
cannot be found. However, by use of methods beyond the scope of 
this book, the values can be computed to as many decimal places as 
desired. The results of such computations have been tabulated m$ 
are included in this text in the form of tables of trigonometric 
functions. Table I at the end of the text contains the values of the 



72 





I 


k r 




Pr V"~ 


-^*(h) 


*\f 


y 


*i 








*i 


V 1 




X 





Jx 



The Trigonometric Functions 
Y 



Sec. 3-4 




Fig. 3-6. 

six functions corresponding to numbers t such that ^ t ^ tt/2. 
Actually, since it/2 = 1.5708 approximately, the table contains 
values of £ between and 1.60. 

Let P(t) = (#, ?/) be the point corresponding to a given value of 
t, Fig. 3-6, and let t x denote the length of the shorter arc which 
joins P(t) to the #-axis. In each case in Fig. 3-6, the point P(ti) 
is located by measuring the arc t x counterclockwise from the posi- 
tive half of the #-axis. We shall call t x the reference number, or 
related number, for t. Note that ^ U = tt/2. Since t x is a real 
number, there is associated with it a point P(t x ) = (x u yi). Also, 
since t x lies between and tt/2, P(£i) must lie in the first quadrant. 

In each case in Fig. 3-6 the coordinates of P(U) must be numer- 
ically equal to those of P{t) ; that is, \x\ — x x and \y\ = yi. Since 
all the trigonometric functions are defined in terms of x and y, we 
can say that 
(3-17) | any function of t \ = same function of t\. 

These functional values may not have the same algebraic sign, 
since P(U) lies in the first quadrant and all functions of U have 
positive values, whereas P(t) may lie in any quadrant and the 
functions of t do not necessarily have positive values. It is impor- 
tant to see that the proper sign is chosen to make the equation a 
true one. The algebraic sign in each case depends on the quadrant 
in which P(t) lies. 

The following examples and Fig. 3-7 will illustrate the method 
of reducing a function of any positive or negative t to the same 
f unctipn of the reference number t x . 

We shall limit our discussion for the present to direct use of 
Table I and consider ipnly values of t x which are shown there. The 
process of using the table for values of t x which are not shown will 
be treated in Section 3-10 when we discuss interpolation. For 
simplicity at this time, we shall use the approximate value 7r = 3.14. 



Sec. 3-4 



The Trigonometric Functions 



73 



t x -W-2 




Example 3-5. Reduce the functions of t = 2 to functions of its reference number. 

Solution: As shown in Fig. 3-7(a), P(2) is in the second quadrant, and the 
reference number ti is it — 2, or 1.14. The numerical values of the functions 
of 1.14 may be found from Table I. The signs of the functions of 2 are determined 
by noting that only the sine and cosecant are positive in the second quadrant. 
Thus, we have : 

cos 2 = - cos (x - 2) = - cos 1.14 = - 0.4176, 
sin 2 = sin (t - 2) = sin 1.14 = 0.9086, 
tan 2 = - tan (tt - 2) = - tan 1.14 = - 2.176, 
cot 2 = - cot (t - 2) = - cot 1.14 = - 0.4596, 
sec 2 = - sec (x - 2) = - sec 1.14 = - 2.395, 
esc 2 = esc (t - 2) = esc 1.14 = 1.101. 

Example 3-6. Find tan 4. 

Solution: As shown in Fig. 3-7(6), the reference number h is 4 — w = .86. 
Since the tangent is positive in the third quadrant, tan 4 = tan .86 = 1.162. 

Example 3-7. Find cos 5. 

Solution: Here, as shown in Fig. 3-7(c), h = 2tt — 5 = 1.28; and cos 5 = 
cos 1.28 = 0.2867. 



Example 3-8. Find cot 20. 

Solution: To locate the point P(20), we must proceed 20 units around the uni$ 
circle in a positive direction from (1, 0). The number of units in one complete 
revolution is 2w = 6.28, find we find that 

20 = 3(6.28) + 1.16. 
Therefore, to locate P(20), we must proceed three times around the unit circle and 
then continue for 1.16 additional units in a counterclockwise direction. This means 
that t may be taken as 1.16. Since P(20) or P(1.16) lies in the first quadrant, 
^ = t = 1.16. From the table, we have cot 20 = cot 1.16 = O.i 



74 



The Trigonometric Functions 



Sec. 3-4 



Example 3-9. Find sin (-2). 

Solution: For <= - 2, it is shown in Fig. 3-8 that ti = | - t - (— 2)| 
= | — tt -f 2 1 = 1.14. Hence, t\ is the same as ^i in Example 3-5. Since sin t is 
negative in the third quadrant, we have 

sin (- 2) = - sin 2 = - sin 1.14 = - 0.9086. 

It should be noted that in Fig. 3-8 the 
points P(—t) and P(t) are located on oppo- 
site sides of the #-axis, and are joined by a 
line segment which is bisected perpendicu- 
larly by the axis. Hence the coordinates of 
the two points are numerically equal, but 
the ^/-coordinates have opposite signs. 
Therefore, by the definitions of the func- 
tions given in Section 3-2, it follows that 




Fig. 3-8. 

sin (— t) = —sin t, 
cos (-£) = cos t, 
tan (— t) = —tan t, 



esc (— £) = —esc t, 
sec (— t) = sec t, 
cot (— t) = — cot*. 



Hence, if t > 0, 

(3-18) (any function of (-*) | = |same function of t\. 

However, the algebraic sign of the function is changed for all 
functions except the cosine and the secant. 

Because cos t and sec t remain unchanged when t is replaced by 
its negative, these functions are called even functions. The remain- 
ing functions are called odd functions, since their values change 
sign when t is replaced by its negative. 



EXERCISE 3-3 

1. Construct a figure, locating each of the following points. Show the point P(t) 
and its related number tu (Use tc = 3.14). 

a. P(l). b. P(3). c. P(10). d. P(- 5). e. P(- 4). f. P(3/2). 

2. With the aid of Table I find each of the following values, using it = 3.14. 
a. sin 1.45. b. cos 3.5. c. sec 4.75. 

d. tan 5. e. esc (- 2.41). f. cot (- 4.50). 

g. sin 28. h. cos 60. i. tan ( - 30). 



. 3ir 
j. sec-£-« 



k. cot 



13tt 



I. sin 



22tt 



3-5. POSITIVE AND NEGATIVE ANGLES AND STANDARD POSITION 

We have defined each of the six trigonometric functions as a 
relation between two sets of numbers, employing as the independent 




Sec. 3-5 The Trigonometric Functions 75 

variable a real number 
t whose absolute value 
represents the length 
of an arc of a unit 
circle. Now we shall 
return to the tradi- 
tional viewpoint and 
consider trigonometric 
functions of angles. 

Although the stu- 
dent is probably famil- 
iar with the idea of 

angle from the study of geometry, we shall try to make the definition 
more precise. Let us select a point in a plane and draw the half- 
line or ray a emanating from O, as shown in Fig. 3-9. We shall 
call O the vertex of the ray. Finally, we let A be a point on the ray 
in its initial position. 

Now rotate the ray a about^O to some terminal position 6, so that 
the point A moves along the arc indicated by the curved arrow AB. 
The ray may be rotated in the counterclockwise sense, as in Fig. 
3-9 (a) , or in the clockwise sense, as in view (6) . Moreover, it may 
be turned through one or more complete revolutions, as in view (c) . 
We shall speak of the position b as the terminal ray b. We have 
then an ordered pair of half -lines consisting of the initial ray a and 
the terminal ray 6. We can now define an angle as follows : 

An angle is a geometric figure consisting of two ordered rays 
emanating from a common vertex. 

With each angle is associated a number, called the measure of the 
angle, which indicates the sense and amount of rotation required 
to turn from the initial ray of the angle to the terminal ray. This 
rotation is usually represented graphically by a curved arrow. Its 
evaluation will be considered in Section 3-6. 

We may designate the angle in Fig. 3-9 as angle AOB; or we may 
use a Greek letter, such as 0, #, a, /3, or y, as the designation. The 
line OA is called the initial side of angle AOB, and OB is the 
terminal side. Counterclockwise rotation, as in Fig. 3-9 (a) oi 
3-9 (c), gives rise tb a positive angle, while clockwise rotation, 
such as the one in Fig. 3-9(6), gives rise to a negative angle. 

Finally, we shall say that an angle is in standard position with 
respect to a rectangular coordinate system when its vertex is at the 
origin and its initial side coincides with the positive ataxia. See 



76 



The Trigonometric Functions 



Sec. 3-5 



i 


\y 




V 






■\ 


^ 


I 









III 




IV 




*»x 



Fig. 3-10. 



Fig. 3-11. 



Fig. 3-10. When an angle is placed in standard position, the 
terminal side determines the quadrant to which an angle is said 
to belong. Thus, angle XOP in Fig. 3-10 is positive because it is 
generated in a counterclockwise direction, and is a second-quadrant 
angle because the terminal side OP lies in the second quadrant. 

We note that the definition of angle does not specify that the 
rotation should stop at the first arrival at the terminal side OP, 
Fig. 3-10. In fact, angles of any size may be generated, since any 
number of angles which end at the terminal side OP of a given 
angle may be obtained simply by adding a number of complete 
rotations, positive or negative, to the given angle. For example, the 
same terminal side may also be reached by rotation in the opposite 
direction. All angles which are in standard position and have the 
same terminal sides are called coterminal angles. 

In Fig. 3-11 the angle a is generated by rotation of OX counter- 
clockwise to the position OP. The angle /3, which is coterminal with 
a, is generated by adding to a one complete rotation of OX. The 
angle y is a negative angle, which is coterminal with a and is 
generated by rotating OX in the clockwise direction to the 
position OP. 



*»x 



3-6. MEASUREMENT OF ANGLES 

The problem of measuring an angle is 
equivalent to that of finding the measure 
of the associated arc. One should, there- 
fore, apply the discussion of Section 3-1 
and construct a unit circle as shown in 
Fig. 3-12. 

Let be an angle in standard position. 
Since the initial and terminal sides of the 
angle intersect the circle in the points P (0) 
Fig. 3-12. an( j P(t), respectively, the problem of 

measuring the angle reduces to that of measuring the appropriate 





$ec. 3—7 The Trigonometric Functions 77 

arc length t Thus, the measure of the angle can be found in terms 
of a real number in any one of several ways, depending on the unit 
of measure chosen. 

We shall consider first the circular system, or natural system, of 
measuring angles, which is used almost exclusively in the calculus 
and its applications. Its fundamental unit is the radian. This unit 
may be defined as follows : 

A radian is the measure of an angle which, if placed at the center 
of a circle, intercepts an arc on the circumference equal in length 
to the radius of the circle. 

In Fig. 3-13 the angle AOB is 
1 radian, and the length of the 
subtended arc AB is equal to the 
radius r. / \ arc =: radius *r 

If the circle selected for meas- 
uring a radian is a unit circle, we 
have an alternate definition of a 
radian. That is, a radian is an 
angle which intercepts a unit arc F IG , 3.43, 

on a unit circle. 

Another system of measuring angles is the sexagesimal system, 
or degree system, which is commonly used in ordinary calculations 
involving angles. The fundamental unit of this system is the degree. 

In Section 3-7 we shall study various relations between radians 
and degrees, and shall develop rules which allow us to convert 
from one system to the other. 

In discussing angles, we frequently use the term angle, in place 
of measure of an angle, and we rely on the context to make the 
meaning clear. Thus, when we say "0 = 2," we mean, "0 is an angle 
whose measure is 2 radians." The word radian is usually omitted 
when an angle is expressed in terms of radians. 



3-7. THE RELATION BETWEEN RADIANS AND DEGREES 

Since an arc that is equal in length to the radius of a circle sub- 
tends an angle of one radian at the center, it follows that the whote 
circumference, which is 2n times the radius, subtends an angle of 
2tt radians. Furthermore, the whole circumference subtends a 
central angle of 360°. Therefore, 

27r radians = 360°, 
and 

ir radians = 180°. 



78 The Trigonometric Functions Sec. 3-7 

If the approximate value 3.1416 is used for 7r, 

180° 180° 



or 
or 
Also, 



1 radian = = t , 1g (approximately), 

7T O.1410 

1 radian = 57.29578° (approximately), 

1 radian = 57°17 / 45 // (approximately). 

1° = r— radians = 0.01745329 radians (approximately). 
loU 



In order to make the conversion to radians easier when the angle 
is expressed in degrees, minutes, and seconds, we give the following 
values : 

1' = 0.00029089 radians, 
and 

1" = 0.00000485 radians. 

Therefore, one of the following rules can be used to convert from 
degrees to radians or from radians to degrees : 

To convert from degrees to radians, multiply the number of 
degrees by ~ , or 0.0174533. 

loU 
To convert from radians to degrees, multiply the number of 

radians by — , or 57.29578. 

Note, however, that certain angles are commonly expressed in 

terms of tt radians, in order to avoid approximate values. For 

example, 

180 = 7r radians, 45 = 7T/4 radians, 

90° = 7r/2 radians, 30° = tt/6 radians. 

3-8. ARC LENGTH AND AREA OF A SECTOR 

In Fig. 3-14 is shown a circle of radius r. In such a circle an 
angle at the center equal to one radian subtends an arc on the 
circumference equal to r. Similarly, by the definition of a radian, 
the number of units in the arc s intercepted by a central angle equal 
to 9 radians is given by the relationship 

.s r 

5 = 1' 
Thus, 
(3-19) $ = rd, 

or 

Fig. 3-14. 
arc = (radius) • (central angle expressed in radians) . 




Sec. 3—8 The Trigonometric Functions 79 

Now let A denote the area of the sector bounded by two radii 
and an arc of length s. If is the number of radians in the central 
angle of the sector, then the ratio of the area A of the sector to the 
area of the whole circle, or 7rr 2 , equals the ratio of the angle to 
the angle in the whole circle, or 2tt. That is, 

— = — > 
7rr 2 27T 

or 

(3-20) A=^r 2 0. 

If the central angle of an arc or a sector is expressed in degrees, 
it must be re-expressed in radians before (3-19) or (3-20) can 
be applied. 

Example 3-10. Express 210° in terms of it radians. 

Solution: Since 1° = ^ radians, 210° = 210 • ~g = -g- radians. 

Thus, 210° = y radians. 

Example 3-11. Express 12°15'20" in radians. 

Solution: Multiply the decimal parts of a radian given in Section 3-7 for 1°, 1', 

and 1" by 12, 15, and 20, respectively. The results are as follows: 

12° 0' 0" = .20943948 radians 

15' 0" = .00436335 radians 

20" = .00009700 radians 

12° 15' 20" = .21389983 radians. 

Example 3-12. Express -^- radians in degrees. 

Solution: Since w radians = 180°, -^- radians = ~ (180°) = 150°. 



Example 3-13. Express 3.5 radians in degrees, minutes, and seconds. 
Solution: First, convert the radians to degrees, as follows: 

3.5 radians = (3.5) (57.29578°) = 200.5352°. 

To find the number of minutes, multiply the decimal part of a degree, or 0.5352, 
by 60. Thus, 0.5352° = (60) (0.5352) minutes = 32.112'. 

To find thfc number of seconds, multiply the decimal part of a minute, or 0.112, 
by 60. The result is 0.112' = (60) (0.112) seconds = 6.72*. 

Hence, 3.5 radians = 200°32'6.72". 



80 The Trigonometric Functions Sec. 3-8. 

Example 3-14. The radius of a circle is 5 inches. Find the length of the arc of 
the circle subtended by a central angle of 30°. 

Solution: Since 30° = -^ > the central angle 6 is -— • Also, r = 5. Therefore, 
by (3-19), 

«=ffl=5-£=5 (3.1416) = 2.618 inches, 
b o 

Example 3-15. In a circle of radius 6 inches, what is the area of a sector whose 
central angle is 60°? 

Solution: By (3-20), the area of the sector is ^r 2 0. Since 6 = 60° = ~ y 

2i 3 

1 7T 

A = o(36)"o~ = 6tt square inches. 





EXERCISE 3-4 






In each problem from 1 to 25, express the given angle in radians. 


1. 60°. 2. 45°. 


3. 30°. 


4. 10°. 


5. 120°. 


6. 150°. 7. 12°. 


8. 90°. 


9. 240°. 


10. 330°. 


11. 72°. 12. 20°. 


13. 215°. 


14. 196°. 


15. 321°. 


16. 283°. 17. 63°. 


18. 30°10'. 


19. 46°21'. 


20. 236°37'. 


21. 82°16'. 22. W2YIT. 


23. 183°57'43". 


24. 392°44'27". 


25. 93°31'38' 



In each problem from 26 to 40, express the given angle in degrees. 
26. tt/6. 27. tt/4. 28. ir/8. 29. 3tt/2. 30. 4tt/5. 

31. tt/12. 32. 5ir/18. 33. 7tt/2. 34. 5x/3. 35. 3tt/20. 

36. 3.7 rad. 37. 8.21 rad. 38. 0.34 rad. 39. 0.763 rad. 40. 0.8136 rad. 

In each problem from 41 to 56, draw the given angle in standard position and 
indicate its terminal side. 

41. 30°. o 42. tt/4. 43. ir/3. 44. 90°. 

45. 22^°- 46. 2t/3. 47. 170°. 48. 17x/18. 

49. - t/2. 50. 630°. 51. 360°. 52. - 4tt. 

53. 7tt/3. 54. 1000°. 55. 9x/4. 56. - 11tt/6. 

57. In a circle of radius 4 feet, find the length of the arc intercepted by an angle of 
7tt/Q radians. Find the angle in radians that intercepts a 5-foot arc. 

58. A central angle in a circle of radius 15 inches intercepts an arc of 5 inches. 
Find the number of radians in the central angle. Express this angle in degrees 
and minutes, rounding off the result to the nearest minute. 

59. A central angle of 62° 14' intercepts an arc of 16 inches on the circumference of 
a circle. Find the radius of the circle. 

60. Find the area of a circular sector whose radius is 7 inches and whose central 
angle is a) 4 radians; b) 75°; c) 3 radians. 

61. The area of a circular sector is 72 square inches. Find the angle if the radius 
is a) 6 inches; b) 9 inches; c) 5 feet. 

62. The area of a circular sector is 126 square inches. Find the radius if the angle 
is a) 128°; 6) 1.6 radians; c) 30°. 



Sec 3-9 The Trigonometric Functions 81 

3-9. TRIGONOMETRIC FUNCTIONS OF ANGLES 

Let be an angle in standard position, as shown in Fig. 3-15. 
With we can associate a real number t, which is the measure of 
the angle in radians. This concept is equivalent to our previous 
concept of t, when t was interpreted as the length of an arc laid off 
on the unit circle by starting at the point (1, 0) and terminating at 




*»x 




P(*,y) 



+x 



Fig. 3-15. 



Fig. 3-16. 



the point P{t). Such an association of the angle with the 
directed length t of an arc of a unit circle allows us to define the 
cosine and sine of 9 as cos = cos t and sin 9 = sin t. A similar 
procedure may be followed for the other functions of 9. 

Now consider Fig. 3-16, where we show an angle 9 in standard 
position and a unit circle. By the definition of 9, the terminal side 
of 9 intersects the unit circle at the point (cos 9, sin 9). This is, of 
course, the point designated previously as P(t). We now extend 
the terminal side of 9 to an arbitrary point P wi th coor dinates 
(x, y). The length of the radius vector OP is r = -\/x 2 + y 2 . 

If we drop perpendiculars from the points (cos 9, sin 0) and 
(x,y) to the #-axis, the right triangles thus constructed are similar. 
Therefore, 

x cos 9 t y __ sin 9 

1 



and - = 
r 



Hence, the coordinates of the point P(x,y) on the terminal side are 

x = r cos and y = r sin 9. 

Using these results with the definitions of the functions from 
Section 3-2, we caii express the values of the six functions in 
terms of x, y, and r. Thus, 

sin 9 = y/r } esc 9 = r/y, 

(3-21) cos 9 = x/r t sec 9 = r/x, 

tan 9 = y/x, cot 9 = x/y. 



82 The Trigonometric Functions Sec. 3-10 

3-10. TABLES OF NATURAL TRIGONOMETRIC FUNCTIONS OF ANGLES 

Tables of natural trigonometric functions are so labeled to dis- 
tinguish them from tables of the logarithms of these functions. 

Angles in Radians. In Section 3-4 Table I was used to find values 
of trigonometric functions of the type cos 2 or sin 27r/3. On the 
basis of the definitions of the functions of an angle given in Section 
3-9, Table I may also be used to find the functions of angles meas- 
ured in radians. 

Example 3-16. Find the cosine of an angle of 1.43 radians. 
Solution: From Table I, cos 1.43 = 0.1403. 

Angles in Degrees. Table II at the end of this text contains the 
approximate values of the six functions of acute angles expressed 
in degrees and minutes. It is a four-place table of the functions of 
angles at intervals of 10 minutes. 

To find the value of a function of an angle between 0° and 45°, 
first locate the angle in one of the columns at the left, and then 
look for the value on the same line in the column headed by the 
name of the desired function. For an angle between 45 p and 90°, 
locate the angle in a column at the right, and then look for the 
value on the same line in the column with the name of the desired 
function at its foot. 

Table II should be referred to in working through these illustra- 
tive examples. 

Example 3-17. Find sin 32°40'. 

Solution: This angle is between 0° and 45°. Look in the left-hand column to find 
32°40', and then go to the right to the column headed sin. There find 0.5398. Hence, 

sin 32°40' = 0.5398. 

Example 3-18. Find cos 56°20'. 

Solution: This angle is between 45° and 90°. So look in the right-hand column 
to find 56°20', noting that 56°20' is above 56°00', and then go to the left to the 
column with cos at its foot. Thus, cos 56°20' = 0.5544. 

The following examples illustrate the procedure for finding an 
angle corresponding to a given value of a function. 



Sec. 3-f-lO The Trigonometric Functions 83 

Example 3-19. Given tan 6 = 4.511, find 0, assuming that 0° £ ^ 90°. 

Solution: Since tan is greater than 1, is greater than 45°. Therefore, search 
through the columns marked tan at the foot for the given number 4.511. The 
corresponding angle in the right-hand column is 77°30'. So 4.511 = tan 77°30', 
or = 77°30'. 

Example 3-20. Given cos = 0.8660, find 0, assuming that 0° ^ £ 90°. 

Solution: By looking through the columns with cos at either the head or the foot, 
find 0.8660. Since this value is in a column headed cos, use the left-hand column for 
the corresponding angle, which is 30°. Hence, = 30°. 

Interpolation. When either the given angle or the given value of 
a function is not printed in the table, we can find the desired value 
or angle by using a method of approximation known as interpola- 
tion. We assume that the change in the value of the function is 
directly proportional to the change in the angle. Although this 
assumption is not strictly valid, it gives values that are accurate 
enough for many practical purposes if we limit its use to small 
changes in the angle. 

The process of direct interpolation is used if the angle is given 
and we need to find the value of either an increasing function of 
the angle, such as the sine, or a decreasing function, such as the 
cosine. Inverse interpolation is used when the value of a trigo- 
nometric function is known and the angle is to be found. 

Example 3-21. Find sin 18°12'. 

Solution: This angle is not listed in the table, but it lies between 18 9 10' and 
18°20'. From the table we find that 

sin 18°10' = 0.3118, 
and 

sin 18°20' = 0.3145. 

The desired value of sin 18°12' will then lie between 0.3118 and 0.3145. 

The tabular difference, that is, the difference between the two values listed in 
the table, is 0.0027. Also, the difference between the angles 18°10' and 18°20' is 
10', while the angle 18°12' differs from 18°10' by 2'. Since the change in the angle 
from 18°10' to 18°12' is 2/10 of the change from 18°10' to 18°20', we assume that 
the corresponding change in the value of the sine will be (0,2) (0.0027) = 0.0005, 
and the amount to be added to 0.3118 is 0.0005. Hence, sin 18°12' = 0.3123. 

The accompanying diagrammatic arrangement presents this same operation in 

tabular form: , . A , 

J sin 18°10' = 0.3118 1 

10 { J sin 18°12' = 0.3118 + x J \ 0.0027 

sin 18°20' = 0.3145 



84 



The Trigonometric Functions 



Sec. 3-10 



Since the angle 18°12' is 2/10 of the way from 18°10' to 18°20', the corresponding 
functional value will be 2/10 of the way from 0.3118 to 0.3145. Therefore, 

x 2_ 

0.0027 10 ' 

x = (0.2) (0.0027) = 0.0005. 

This amount is to be added to 0.3118. Hence, the value of the function is 0.3123, 
and sin 18°12' = 0.3123. 



or 



Example 3-22. Find cos 73°48'. 

Solution: The process is similar to that in Example 3-21. However, since the 
cosine decreases as the angle increases, we subtract 8/10 of the tabular difference 
from cos 73°40'. We find the values of cos 73°40' and cos 73°50' in a column of the 
table labeled cos at the bottom. The work may be indicated as follows: 

cos 73°40 ; = 0.2812 

8 { } x 

cos 73°48' = 0.2812 - x 



10 



cos 73°50' = 0.2784 



0.0028 



g^g = ft ' or x = (°* 8) (0 - 0028) = °- 0022 - 
Hence, the amount to be subtracted from 0.2812 is 0.0022, and cos 73°48' = 0.2790, 

The inverse process of finding the angle when the given value of 
a function is not printed in the table is performed in a similar 
fashion. Here, since we know the value of the function, we find 
the two values in the table nearest the given value, one less than 
it and one greater. Again making the assumption that small 
changes in the value of the function are proportional to small 
changes in the angle, we proceed as indicated in the following 
example. 



Example 3-23. Find if cot = 0.8780. 

Solution: This value of the cotangent is not in the table but lies between the 
entries 0.8796 and 0.8744. To these correspond, respectively, the angles 48°40' 
and 48°50'. We have, therefore, the following tabulation: 

f cot 48°40' = 0.8796 1 

x \ 0.0016 

10 { [ cot 48° (40 + x)' = 0.8780 J \ 0.0052 



cot 48°50' 



= 0.8744 



Hence, = 48°43 / . 



* 16 A 
10 = 52 ,and * =3 ' 



Sec. 3-10 The Trigonometric Functions 85 

EXERCISE 3-5 

In each of the problems from 1 to 30, use Table II to find the value of the given 
function. Interpolate whenever necessary. 

1. sin 36°20'. 2. cot 12S°40'. 3. sec 23°40'. 

4. cos 96°50'. 5. sin 132°10'. 6. tan (- 2S°10'). 

7. esc 223°30'. 8. sec 39°30'. 9. cot 283°50'. 

10. sin 98°40'. 11. cos 75°C0'. 12. cot (- 133°30'). 

13. sec (- 392°10'). 14. esc (- 416°20'). 15. tan G23°40 / . 

16. tan 298°52'. 17. cot 55°43'. 18. esc (- 44°51'). 

19. sin 57°32'. 20. cot 3°1G'. 21. sin (- 280°33'). 

22. cot 28°01'. 23. tan 27°16'. 24. esc (- 245°29'). 

25. cos (- 72°58'). 26. sin 312°37'. 27. tan 636°02'. 

28. sin (- 16°47'). 29. esc 2S9°0G / . * 30. cos 12G°19'. 

In each of the problems from 31 to 60, use Table II to f.nd the values of between 
0° and 360° which satisfy the given equation. Express the results to the nearest 
minute, interpolating whenever necessary. 
31. tan = - 0.11CS. 
34. cos0 =0.7951. 
37. sin = 0.5783. 
40. cos = - 0.4147. 
43. tan = 8.345. 
46. sin = 0.G702 
49. ccs = 0.9503. 
52. cot = - 1.381. 
55. cot = 7.C00. 
58. cot = 0.1340. 

In each of the problems irom Gl to 72, find the value of the given function. 
Interpolate whenever necessary. Take t as 3.14. 

61. sin 0.93. (7. cot 2.46. 63. sec (- 1.24). 

64. tan 8.71. C5. esc 9.43. 66. cot 0.G78. 

3tt 
4 ' 



32. 


sin 6 = 0.30G2. 


33. 


cot 6 = 1.091. 


35. 


tan 6 = 0.0553. 


36. 


sin 6 = 0.2419. 


38. 


cot 6 = - 0.G494. 


39. 


tan 6 = 1.511. 


41. 


sin~0 = - 0.9959. 


42. 


cot = 0.0437. 


44. 


cot 6 = - 0.3121. 


45. 


tan = 1.446. 


47. 


tan 6 = 0.9043. 


48. 


cot 6 = 2.398. 


50. 


cos 6 = - 0.5090. 


51. 


cos 6 = 0.S519. 


53. 


cot 6 = 0.4230. 


54. 


sin 6 = 0.2491. 


56. 


tan = -0.1191. 


57. 


ccs 6 = - 0.1323. 


59. 


tan 6 = - LS.C0. 


60. 


tan = 3.235. 



67. tan 0.333. 68. cot ( - t) 69. cos 



70. sin 



( - ~) • 71. esc 0.968. 72. cot ( - 0.643). 



In each of the problems from 73 to 84, find the values of 6, in radians, between 
and 27T which satisfy the given equation. Use Table I and express the results to 
three decimal places, interpolating whenever necessary. 

73. cos 6 = 0.9759. 74. sin 6 = 0.99G7. 75. tan = 2.066. 

76. sec = 2.563. 77. tan = 0.9413. 78. sin = - 0.7360S 

79. cos = 0.4010. 80. tan = l.G. 81. sin = 0.91. 

82. cot 6 = 0.39. 83. cos = 0.84. 84. cot = 1.031. 



4 



The Laws of Exponents 



4-1. POSITIVE INTEGRAL EXPONENTS 

When studying the progress of algebra up to the sixteenth 
century, one cannot help but be perplexed by either the total 
absence of symbolism or, when present, the lack of uniformity in 
its use. At first, unknown quantities were often represented by 
words. Later, symbols made from abbreviations and initial letters 
of these words were used to indicate mathematical concepts, such 
as number, power, and square. 

Descartes (1637) is generally credited with our present system 
of exponents. He introduced the Hindu-Arabic numerals as expo- 
nents, using the notations a, aa [sic] , a 3 , a 4 , etc. The writing of a 
repeated letter for the second power of the unknown continued for 
many years. 

Laws for positive integral exponents were introduced in Section 
1-11, without proofs. We shall now establish these laws and extend 
them to apply also to zero, negative, and fractional exponents. 

We recall that if n is any positive integer, a n means the product 
of n factors each equal to a. In this notation, a is the base and n is 
the exponent or power. We shall proceed to establish the following 
laws for positive integral exponents. 

Law of Multiplication. If a is a real number, and if m and n are 
positive integers, 
(4-1) a m • a n = a m+n . 

Proof. Proof of this relationship follows from the definition of a n 
and the associative law for multiplication. Thus, 

a m = a • a a (tow factors), 



and 



Hence, 



a n = a * a a (ton factors). 



a m • a n = [a • a • • • • a(to m factors)] [a* a a(to n factors)] 

= a • a cr(to m + n factors) 



= a 



m+n 



For example, x s • x 5 = x 8 , and y k • y k+3 = y 2k + 3 . 

86 



Sec. 4—1 The Laws of Exponents 87 

Law of Division. If a is a non-zero real number, and if m and n 
are positive integers such that m > n, then 

(4-2) Z- = a™~\ 

If a ¥= 0, and if n > m, then 

n m 1 

(4-3) ^L = -J_ . 

v y a n a n - m 

Proof. Proofs of these relationships follow : 
If m > n, then m — n is positive. By (4-1) , 

a m-n . a n — a (m-n)+n — a m # 

Hence, dividing both sides by a n , we have 

a m 
a m ~ n = — • 
a n 
For example, 

-t = x 3 j and 7^ = 3 2 . 
If m < n, then n — m is positive. By (4-1), 

Divide both sides by a n ~ m to obtain 

a n 

a m = 



a n-m 

Now, dividing both sides by a n , we have 

a m ( a n \ / a n 1 1 

a n \a n - m J / a n a n ~ m a n ~ m 

For example, 

z 4 1 , 3 5 1 

5*"?' and 3* = 3*' 

Law for a Power of a Power. If a is a real number, and if m and 
n are positive integers, then 

(4-4) (a m ) n = a mn . 

Proof. This relationship can be easily proved as follows : 

By the associative laws for multiplication and addition, the law 

of multiplication expressed by (4-1) can be extended to three or 

more factors. Thus, 

a m • a n • a p = (a m • a n ) • a p 

= a m+n • a p 

A similar relationship can be written for any number of factors. 
That is, 

(a m ) (a*) • • • (a r ) = a m+ *+ •••+'. 



88 The Laws of Exponents Sec. 4-1 

We may now take m = p = • • • = r to get 

(a m ) (a m ) • • • (a w ) (to n factors) = a w + w+ •••+» = a»»n # 
Hence, 

(a w ) n = a mn . 
For example, 

(a; 2 ) 3 = x 6 , and (2 2 ** 1 ) 5 = 2 10A + 5 . 

Law for a Power of a Product. If a and 6 are real numbers, 
and if m and n are positive integers, then 
(4-5) (ab) n = a n b n . 

Proof. In proving this relationship, we make use of the associa- 
tive and commutative laws of multiplication. Thus, 

{ab) n = (ab) • (ab) (ab) (to n factors) 

= [a * a a(to n factors)] [b • b 6(to n factors)] 

= a n b n . 

For example, 

(- 2xV) 6 = (~ 2) 5 s 1 V 6 = - 32z 1 V 5 . 

Law for a Power of a Quotient. If a and b are real numbers, if 
6 ¥" 0, and if n is a positive integer, then 

Proof. By applying the law for multiplying fractions, we have 
W =6*6 ^ (ton factors) =^. 

For example, 

(3xy k \ 2 _3 2 x 2 y 2k _ 9s 2 y 2 * 
V z 2 / " z 4 ~ z* ' 

An exponent affects only that quantity to which it is attached. 
Thus, ~5x(y 3 ) 2 = — 5xy Q , whereas (— 5xy 3 ) 2 = 25x 2 y Q . 

So far we have defined a n only when n is a positive integer. We 
shall now introduce zero, negative integer* and rational powers in 
such a way that they will obey the same laws which were proved 
for positive integral exponents. 

4-2. MEANING OF a 

We shall define the zero exponent by the equation 
(4-7) a° = 1 (a^ 0). 

A few illustrations are : 

aP = l, 7° = 1, 5(2zt/*)o = 5, (a» - 6*)° = 1, (^)° = 1. 



Sec. 4-2 The Laws of Exponents 89 

If a in (4-2) is not zero and m = n, we get 

a m 

— = a m ~ n = a . 

a n 

In this case, the quotient on the left equals 1, while the value of the 
term on the right is a . Since a = 1, by definition, the law of 
division holds for n = m, as well as for m> n and n> m. 

The student should note that (4-7) gives the only possible defini- 
tion of a if the law expressed by (4-2) and (4-3) is to hold for the 
zero exponent, as can be seen from the foregoing discussion. 

We shall show that the definition a = 1 is consistent with the 
five laws of exponents in Section 4-1 ; that is, we shall show that 
these laws also hold when any exponent is zero. In the following 
explanations, where a quantity occurs in a denominator, we assume 
that it is not zero. Also, the exponents are assumed to be non- 
negative integers. 

Let us, for sake of discussion, suppose that n = in (4-1), that 

1S > 1X1 a m . a n = a m+n m 

Then we have n m 1 

a m • a n = a m • a = a m • 1 = a m , 

or 

Hence, the law of multiplication holds when n = 0. A similar 

procedure will verify the law if m = 0. 

Now let us suppose that n = in (4-2), that is, in 

a m 

— = a m ~ n . 

Then , 

Qin Qin Qta 

— = — = —- = a m , and a m ~ n = a m ~° = a w . 
a n a 1 ' 

Hence, (4-2) holds when n = 0. 
If m = in (4-3) , we have 

£!! - ^ - J_ , d 1 1 = 1 t 

a n ~" a n ~" a n ' a n ~ m ~~ a n_0 ~" a n 

It is clear that in (4-2) m cannot be zero, and in (4-3) n cannot 
be zero. Therefore, the law of division holds. 
Suppose that n = in (4-4) , that is, in 

(a m ) n = a mn . 
Then 

(a m ) n =i (a w )° = 1, and a mn = a w, ° = a = 1. 

If m = in (4-4) , we have 

(a™) n = (a°) n = l n = 1, and a mn = a°' n = a = 1. 
Hence, the law for a power of a power holds. 



90 The Laws of Exponents Sec. 4-2 

Now consider (4-5) , which is 

(db) n = a*b n . 

If n = 0, 

(ab)» = (ab)° = 1, and a»b n = a°b° = 1 • 1 = 1. 

Finally, we let n = in (4-6), that is, in 

(ay_ <r 

W ~ b n ' 
Then , ffi> - - o . a » a o j 

y = w =i ' and t«=¥ = i =i - 

The demonstrations just given prove that the five laws of expo- 
nents, originally stated for positive integral powers, are true for 
all non-negative integral powers, and that the law of division is 
true even when the exponents are equal. 

4-3. NEGATIVE EXPONENTS 

In order to extend the meaning of exponents to negative integers, 
we define ar n by the following relationship : 

(4-8) or- = ± {a* 0), 

where n is a positive integer. 
Several illustrations are: 

5 - 2= ^=4' 1^ = 103 = 1000, 

(a - to) ^ = (T^Eo 5 ' and @ 1= i = i' 

y 

As in Section 4-2, we shall show that our definition is consistent 
with the five laws of exponents. 

Let us first note that (4-8) is true even if n = or if n is a nega- 
tive integer. If n = 0, then 

a -n = a o = ! = 1 J_ = 1 . 
1 o° a n 

If n = — p, where p is a positive integer, then a~ n = a p = 1 /— 

~~ er p ~~ a n # 

We shall use this result in the proofs that follow. 

In order to extend (4-1), let m be a non-negative integer, and let 
n be a negative integer, say, w = —p. In this case it is also assumed 
that a ¥> 0. Then 

1 a m 
a m • a n = a m • a~ p = a m • — = — • 

a p a p 



Sec, 4—3 The Laws of Exponents 91 

If m ^ p, we have, by (4-2) for non-negative exponents, 

d m 
a m • a n = — = a m ~ p = a m+( ~ p) = a m+n . 
a p 

If m<p, we have, by (4-3) for non-negative exponents, 

o m 1 

a m • a w = — = = a~ (p ~ m) = a w ~ p = a m+n . 

A similar demonstration establishes (4-1) in case n ^ and 
m < 0, or in case ra < 0, and n < 0. 

Proof of the extension of (4-2) rests on the validity of the law 

of multiplication just established. If a¥=0, and if m and n are 

integers (positive, negative, or zero), then 

a m 1 

— = a m — = a m • ar n = a w_n . 

a n a n 

Although (4-3) is now an immediate consequence, it is not really 
needed, in view of the general validity of (4-8). The demonstra- 
tion just given allows the law of division to be stated as a single 
relationship as follows : 

(4-9) *— = a m ~ n (a^O). 

Thus, a single law applies, regardless of whether m > n, n > m, or 
m = n, where m and n are arbitrary integers. 

Now consider the law for a power of a power. In (4-4) let 
n — —p, where p^O, while m ^ 0. Then 

K a ) - w ) - (am)p amp 

Also, 1 

a mn = a m (~ p ^ = a —,np = • 

Hence, (4-4) holds in this case. 
If m ~ —p y where p ^ 0, while n S 0, we have 

/ 1 \ w 1 1 

(a™)" = (ar p ) n = ( — ) = — i and a mn = a ( ~ p)n = a~ pn = — • 
v J K J \a p / a np a np 

Again (4-4) holds. If both m and n are negative, a similar proce- 
dure is used, and the extension of (4-4) holds. 

To extend the law for a power of a product, let n = —p in (4-5), 
where p ^ 0. Then 

(a&) n = (a6)~ p = y-^- = —r- > and a n b n = a- p 6~ p = — . — = — -$ 
v ' v ' (a6) p a p b p a p b p a p b p 

So (4-5) is verified for negative integral values of n. 

To verify the exltended law for a power of a quotient, assume 
that n = — p in (4-6), where p = 0. Then 



W "" W "" /ay ~ /6* " 



6 p , a n cr* 6 P 

-— > ana tz =r 7~r = -z 
a? b n * b~ p o? 



92 The Laws of Exponents Sec. 4-3 

Hence, (4-6) holds for negative integral values of n. 

Thus, the laws of exponents hold for positive integral exponents, 
zero exponents, and negative integral exponents. In Section 4-5 
we shall consider the case of fractional exponents. 

From the general validity of (4-8), it follows immediately that 
a factor of the numerator or the denominator of a fraction can be 
moved from the numerator to the denominator, or vice versa, pro- 
vided only that we change the sign of its exponent. For example, 

a*b~° = tq > and 



b 3 yz~ 2 y 



SCIENTIFIC NOTATION 



We are now in a position to introduce certain simplifications 
when operating with very large or very small numbers, as are 
customarily used in scientific writing. Any positive number that is 
greater than 10 or less than 1 may be written compactly by 
expressing it in standard form, that is, by writing it as a number 
that lies between 1 and 10 multiplied by a suitable positive or nega- 
tive integral power of 10. Thus, 27,000 would be written 2.7 • 10 4 . 
Similarly, 0.00031 would be 3.1 • 10~ 4 . 

Example 4-1. The speed of light is 186,000 miles per second. Express this 
number in scientific notation. 

Solution: The given number 186,000 may be written as 1.86 • 10 5 . 

Example 4-2. The rest mass of an electron is 9.11 • 10~ 28 grams. How many 
zeros would be required between the decimal point and the first non-zero digit, 9, if 
the number were written in decimal notation? 

Solution: The exponent — 28 means that we would have to move the decimal 
point 28 places to the left from its present position. We would thus have to place 
27 zeros to the left of the 9. 

Example 4-3. If the sun is 9.3 • 10 7 miles from the earth, how long does it take 
light to reach the earth from the sun? 

Solution: As given in Example 4-1, the speed of light is 1.86* 10 5 miles per 

9 3 • 10 7 
second. Therefore, the required time is z ■ -' ; ^ = 5 • 10 2 = 500 seconds = 

8 minutes 20 seconds. 



4-5. RATIONAL EXPONENTS 

We shall now extend the meaning of exponents from integers to 
rational numbers. Here again we shall make the extension in such 



Sec. 4-5 The Laws of Exponents 93 

a way that the laws for positive integral exponents will be 
preserved. 

Suppose that a is a real number and that n is a positive integer. 
Let us assume that a 1/n has meaning and that (4-4) applies. Then 
it would be true that 
(4-10) {a l ' n ) n = a^ ,n)n = a 1 = a. 

This says that the nth power of a Vn would have to be a, or in other 
words that a 1/n would be what is called an nth root of a. For 
example, (4-10) would yield 

(a 1 ' 2 ) 2 = a, and (a 1 / 3 ) 3 = a. 

Real nth Roots of a. Before defining a 1/n , let us examine the sit- 
uation with respect to the existence of nth roots of a given number 
a. The following results may be proved with the help of the theory 
of equations. 

Case I. If n is an even integer and a is a positive real number, 
there are two real numbers that satisfy the equation r n = a. One 
of these is the positive nth root of a, which is denoted by tya. The 
other is the negative nth root of a, which is denoted by — tya. We 
may also denote these two numbers together by zt^J/a. 

Case II. If n is an even integer and a is a negative real number, 
no real nth roots exist, since no even power of a real number can 
be negative. 

Case III. If n is an odd integer and a is a positive real number, 
there is one real (positive) value of r such that r n = a. In other 
words, if n is odd, there is a real positive nth root, which is denoted 
by tya. 

Case IV. If n is an odd integer and a is a negative real number, 
there exists one real (negative) value of r such that r n — a. That is, 
if n is odd, there is a real negative nth root, which is denoted by 

Case V. If n is any positive integer and a is zero, there is only 
one real nth root, and this root is zero. 

Thus, the definition of an nth root of a is valid under all condi- 
tions except when a is negative and n is even. In this situation, no 
real nth roots exist. (However, the introduction of complex num- 
bers in Chapter 11 will allow us to eliminate this exception.) We are 
now ready for the following definition. * 

Definition. If a is a non-negative real number and n is a positive 
integer, a Vn designates the non-negative nth root of a, or tya. If a 
is negative and n is an odd positive integer, then a 1/n designates the 
real nth root of a, or ^/"a. 

When a is negative and n is even, a 1/n is undefined: 



94 The Laws of Exponents Sec. 4-5 

Meaning of a m/n . Let a be a given real number, n a positive 
integer, and m an integer. If a m/n has meaning, and if (4-4) holds 
for fractional powers, then a m/n = a (1/n)TO = (a l/n ) m . Under these 
assumptions, then, a m/n would be the mth power of a 1/n . It is 
natural to state the following definition. 

Definition* If n is a positive integer, if m is any integer such that 
the fraction m/n is in lowest terms, and if a is a real number which 
is assumed to be non-negative when n is even, then a m/n designates 
the mth power of a 1/n , that is, the mth power of tfa. Hence, 
(4-11) a mln = (a 1/n ) m . 

If the fraction m/n is not in its lowest terms, it is first reduced to 
lowest terms, and (4-11) is then applied. 

When a is given, the value of a m/n depends only on the value of 
the fractional exponent, not on the particular values of m and n. 
Thus, 

2 4/2 = 2 2 = 4, 2** = 2 3 ' 4 , and (- 2) 2 ' 6 = (- 2) 1 ' 3 = JT^2. 

In the last example it would be incorrect to apply (4-11) directly, 
since (— 2) 1/6 has no meaning. 

It may be shown that, if a is positive, 

(4-12) a mfn = *s/cr. 

The proof is omitted. 

We shall also omit the details of the procedure for showing that 
the five laws of exponents hold for rational exponents and non- 
negative bases. The reader is cautioned against using the laws for 
negative bases, since some fail under certain conditions. 

To summarize the results now established, we restate the laws of 
exponents here for easy reference. It is assumed that a and b are 
non-negative real numbers, and that m and n are rational numbers. 
Furthermore, if either a or 6 appears in a denominator or raised 
to a negative or zero power, it is assumed to be different from zero. 

Law of multiplication: a m * a n = a m+n . 

d m 
Law of division: — = a m ~ n . 
a n 

Law for a power of a power: (a m ) n = a wn . 

Law for a power of a product: (ab) n = a n b n . 

(a\ n a n 
h) ~ fe~ # 

Law for reciprocal: a~ n = ~ • 
Zero power: a = 1. 



Sec. 4-5 The Laws of Exponents 95 

Note that the radical notation can be replaced by the simpler 
and much more convenient exponential form. Everything that can 
be done with the radical notation in the simplification of roots of 
numbers and in operations involving roots can be done much more 
naturally by means of the exponential notation. A few illustrations 
of the meanings and uses of exponential forms follow : 

x 1 ' 2 = y/i, if x ^ 0; (- 27) 1 ' 3 = ^/^27 = - 3; 
- (32) 1 ' 5 = - S/32 = - 2; - ^fi = - a 2 ; 
32 2 ' 5 = (^32) 2 = 2 2 = 4; - (16a; 4 )- 1 / 2 = " * "~ * "" l 



(16a; 4 ) 1 / 2 y/Tfat 4x 2 
The following examples illustrate the applications of the laws of 
exponents to the solution of problems involving radicals. 

Example 4-4. Compute the value of \/2 • y/% and write the result in expo- 
nential form. 

Solution: Using exponential notation and the laws of exponents, we have 

2i/3 . 21/4 = 24/u? . 2 3 ' 12 = 2 4/12+3/12 = 2 7/12 . 



Example 4-5. Remove all possible factors from the radical ^162x 4 2/ 2 . 
Solution: We may proceed as follows: 



^162z 4 z/ 2 = (2 • 3 4 *V) 1/8 = 2 1 ' 3 • 3 4 ' 3 .r 4 ' 3 ?/ 2 ' 3 
= 3x • 2 1 ' 3 3 1/3 z 1/3 2/ 2 ' 3 = 3x VOxy*. 

Example 4-6. Use the laws of exponents to express y/x • *tyy by using only 
one radical. 

Solution: Changing to fractional exponents, we have 

y/x* y/y = x ll2 y 113 = z 3 'V /6 = (» 8 2/ 8 ) 1/a = \/» a V 2 - 

Example 4-7. Rationalize the denominator in the fraction ~^== • 

Solution: To write an equivalent fraction in which no radical appears in the 
denominator, we proceed as follows: __ 

\_ 1 _ 1 s 2/5 _ s 2/5 y/&_ # 
^/x z ~~ X 315 ~~ X 315 * x 2 ' 5 ~ x ~~ x 

Example 4-8. Ratiorialize the denominator of the fraction -= • 

5 - V3 

Solution: We use the relationship (a +b) (a — b) = a 2 — b 2 to remove the 
radical from the denominator. Thus, 

1 1 . 8 + \/ 8 = 5 + V 3 - 5 ± V 3 - 

5 - V 3 5.- V3 5 + V3 25-3 22 



96 The Laws of Exponents Sec. 4-5 

x 2 



<\A 2 - x 2 + ■ 



*Aj2 #2 

Example 4-9. Change _ \ to a simple fraction. 

Solution: Write the expression in exponential form, as follows: 

<* -«■>"' + («..!%.>. 

a 2 — a; 2 

Then, multiplying the main numerator and the main denominator by (a 2 - x 2 ) 112 , 

we have „ 9 , 9 9 

a 2 - .r 2 + x 2 a 2 



(a 2 - x 2 ) 3 ' 2 (a 2 - x 2 ) 3f2 

Example 4-10. Express (x 2 + a 2 ) 3 ' 2 + 3z 2 (x 2 + a 2 ) 1 ' 2 in a factored form. 
Solution: Rewrite the expression as 

{x 2 + a 2 ) 112 (x 2 + a 2 ) + Sx 2 (x 2 + a 2 ) 112 . 
Removing the common factor (x 2 + a 2 ) 112 , we obtain 

(x 2 + a 2 ) 112 [x 2 + a 2 -j- 3a; 2 ] = (x 2 + a 2 ) 1 / 2 (4a; 2 + a 2 ). 

EXERCISE 4-1 

In each of the problems from 1 to 20, perform the indicated operations and 
eliminate all zero and negative exponents. 

1. 3z°y. 2. ar 1 ' 2 . 3. (|)~ 3 - 4. 10- 2 . 

5. j~zr 6. (|)°- 7. (9*)i/». 8. faV«-». 

9. (»•)-»'*. 10. fo.V'i, • 11- ^f^ • 12. 3s W • ±x" 2 yz\ 

(a? l/4 2/) 4 a 2 t/" 9 y ^ 

13. (a;V)*(^"V 4 ) 2 . 14. (& l 'V a ) 4 (s a y" 1/2 )°- 15- (* 1/8 H-y l/a ) 2 . 

16. ar* + y x . 17. (a; + y)" 1 . 18. (x + y)' lf3 (x + 2/) 1/3 . 

19 ! - *" 1/2 . 20 * 1/2 + 3" 1/2 , 

1 + ar 1 ' 2 ' * a: 1 ' 2 -a;- 1 / 2 ' 

Write each of the following expressions in exponential form. Remove all possible 
factors from the radical and, wherever necessary, rationalize the denominator. 

21. V§6. 22. ^~=~54. 23. ^2l. 24. y/xy*. 

25. tfflyi. 26. </Fi. 27. v/l2r=^. 28. ^"^. 



29. ^(-64)°y-». 30. Wx°y-»)*. 31. V^ly" 2 )" 1 . 32. vV + b 2 )- 1 . 
33. V3 • ^14. 34. -#? • V5- 35. V^ v^. 36. #ai&5 - 4 /8^. 
37. — 5_ = . 38. —J 39. — 2_ . 40. X 



3 - V2 V6 + 2 1 - V5 V3 - V2 



41. i_. « B 5±3£ I =». tt l+ ^^ u.*£=J^. 

3-ya: 2 «-9 3-vz 2 -9 z-y/z*-y* Vz+V^+l 



Sec. 4-7 The Laws of Exponents 97 



Vl - X 2 + —~= 

x(l - x) Aa y/\ - x* 



45. V2x - x 2 + * v . .JL . 46, 



1 + Vl + * 2 - 



y/2x - x 2 \ - x 2 

x 2 



47. __ — -^^jQ+^L . 48. ( x a + 1)3/2 + X 2( X 2 + 1)1/2. 

(1 + VI + * 2 ) 2 
49. (2 - x 2 ) 5/2 + .r 2 (2 - x 2 )* 12 . 50. (a* - 3) I/2 - z 2 (z 2 - 3)" 1 ' 2 . 

4-6. THE FACTORIAL SYMBOL 

The product of all positive integers from 1 to n inclusive is called 
"n factorial" or "factorial n" and is represented by either of the 
symbols n ! or An. Thus, if n is a positive integer, 

n! = 1 -2-3 (n - 1) • n. 

For example, 

3! = 1-2-3 = 6; 5! = 1-2- 3-4- 5 = 120; 

6! = 5!- 6; j = 7!; r! = [(r - l)!]r. 

4-7. THE BINOMIAL THEOREM 

The statement known as the binomial theorem enables us to 
express any power of a binomial as a sum of terms without per- 
forming the multiplications. 

By actually performing the indicated multiplications, we find that 

(a + 6) 2 = a 2 + 2ab + 6 2 , 

(a + b) z = a 3 + 3a 2 b + Sab 2 + fc 3 , 

(a + b) 4 = a 4 + 4a 3 6 + 6a 2 b 2 + 4afe 3 + ft 4 . 

These formulas may be rewritten in the following manner, so as 
to suggest a general rule 1 : 

(a + b) 2 = a 2 + \ab + < ~b 2 i 

(a + 6) 3 = a 3 +\a% + f-fjafc* + f-ff^ 3 , 

(a + ^ = a* + fa 3 6 + *f§«V + f^fafc 3 + f^^*. 

Applying this suggested rule to (a + 6) 5 , we obtain 
(a + &)« = as +^ a«6 + £l| a»6» + ^|t| "^ 

, 5-4-3-2 fc4 , 5 « 4 » 3 • 2 • 1 ,,, 

+ 1.2.3.4 qb4+ l-2.3.4.5 b - 

1 The justification for writing the expressions on the right in this form will 
be found in Chapter 17, where the binomial coefficients are 'given in terms of 
the combination formulas. 



98 The Laws of Exponents Sec. 4-7 

Upon simplification of coefficients, we get (<x + b) 5 = a 5 + 5a 4 b + 
10a 3 6 2 + 10a 2 6 3 -I- 5a6 4 + 6 5 , which is the same result as that 
obtained by multiplying (a + &) 4 by (a+6). 

Each of the expressions on the left is of the form (a + b) n , in 
which the exponents 2, 3, and 4 of (a + b) are special values of n. 
If we let n denote the exponent of (a + b) in each of the expres- 
sions on the left, we note that the expansion of (a + b) n contains 
n + 1 terms with the following properties : 

1. In any term the sum of the exponents of a and b is n. Also, 
the first term is a n and the last term is b n . 

2. The exponent of a decreases by 1, and the exponent of 6 
increases by 1, from term to term. 

3. The denominator of the coefficient in each term is the fac- 
torial of the exponent of b in that term. 

4. The numerator of the coefficient in each term has the same 
number of factors as the denominator. Specifically, wherever 1 
appears in the denominator, write n directly above it in the 
numerator ; wherever 2 appears in the denominator, write n — 1 
directly above it in the numerator; and so on. Thus, in (a + b) 5 , 
the number above 1 is 5, and the number above 2 is 4. 

Assuming that these properties hold for all positive integral 
values of n, we have 

(4-13) (a + b) n = a» + j a^b + nf * ~ 1} a- 2 6 2 

, n(n — 1) (n — 2) n „, a . , , „ 

+ — — 1.2*3 + * " ' + 

This result is the binomial formula. So far we have verified this 
formula only for n = 2, 3, 4, and 5. In Chapter 16, we shall prove 
the binomial theorem, which states that the formula is true for all 
positive integral values of n. 

The following example shows the procedure for the expansion of 
(a + 6)\ 

Example 4-11. Expand (x - 2?/) 6 by the binomial theorem. 

Solution: The required expansion will be obtained by letting a = x, b = — 2y, 
and n = 6. We begin by setting up the following pattern of n + 1 terms : 

x 6 + s*( - 2y) + x*{ - 2y) 2 + x 3 ( - 2?/) 3 

+ x*( - 2yY + x( - 2yY + ( - 2y)\ 



Sec. 4-8 The Laws of Exponents 99 

The exponents of b in the second, third, fourth, and fifth terms are 1, 2, 3, 4, and 5, 
respectively. Hence, remembering that the last term is b n , we may fill in the 
numerators and denominators of the coefficients as follows: 

*• + j *•(- 2y) + 5l| *«(- 2»)» + |f|f| «•(- 2»)« + \\\'X.\ **{- W 

By simplifying, we obtain the following result: 
(x - 22/) 6 = z 6 - 12x 5 2/ + 60s 4 i/ 2 - 160z 3 2/ 3 + 240x 2 */ 4 - 192xy* + 64y 6 . 

4-8. GENERAL TERM IN THE BINOMIAL EXPANSION 

If we wish to write any particular term of the expansion of 
(a + b) n without considering any of the other terms, a study of the 
binomial formula in (4-13) Section 4-7 will reveal the' following 
facts : 

In every term the exponent of 6 is one less than the number of 
the term. Thus, in the (r+ l)th term, the exponent of b is r. 
(The expression for a particular term is simplified slightly if the 
number of that term is called r + 1, rather than r.) 

The sum of the exponents of a and 6 is n in each term. For 
the (r 4- l)th term the exponent of a is n — r. 

The denominator of the coefficient in the (r + l)th term is r!, 
since it is the factorial of the exponent of b. 

The numerator of the coefficient has the same number of fac- 
tors as the denominator. In the (r + l)th term, it is the product 
n(n — 1) (n — 2) • • • (n — r + 1). 

We obtain, then, for the (r + l)th term of the expansion 
(« + &)», 

n(n - 1) (n - 2) ■ ■ ■ (n - r + 1) alMftr 
r! 

Example 4-12. Find jihe sixth term of (3x - y 2 )*. 

Solution: Here a = 3#, b = - y 2 , and n = 8. Since r -f 1 = 6, the exponent of 

b is r = 5. Hence, the exponent of a is n - 5 = 3. Therefore, the sixth term is 

8-7-6-S-4 



1*2<3*4*5 



(3x)H-V 2 ) 5 = -15123V . 



1 00 The Laws of Exponents Sec. 4-8 

EXERCISE 4-2 

In each of the problems from 1 to 16, reduce the given fraction to lowest terms. 



1. 
5. 
9. 



(91) (31) , 
8! 

(3-4)!, 
3(4!) 

n! 
(n-2)!' 



2. 

6. 

10. 



10! 



(3!) (7!) 
(3-4)! 



(3!) (4!) 
n! 



13. 
15. 



(n - r) ! 
(» + l)!(n -1)! 



(n!)» 
n!(n + l)l 



3. 

7. 

11. 



(10(3!) , 
(2!) (2!) 

3! +4! 
(3!) (4!) * 

(n + 1)! 
(n-1)!' 

[(n + 1)!] 2 
n\(n -2)! 

n\(n -2)! 



4. 

8. 

12. 



(7!) (8!) , 
(5!) (6!) 

nl 

(n-1)!" 

n! 
3!(»-3)!' 



17. Show that 



(n -l)!(n+2)l 

n(n - 1) (n - 2) 



14. 
16. 



[(n-l)l]» 
(n-r + 1) _ 



n! 



r! 



r! (w — r)! 



In each of the problems from 18 to 32, expand the given expression by the 
binomial theorem. (Hint: In problems 29 thru 32, first consider the first two terms 
in parentheses as a single quantity). 

18. (x + y): 19. (* - 1) T . 20. (a - 26)*. 

23.(yx+-tY - fx 2>6 



21. (2a* - 362)'. 

25.P-4)\ 

\y x 2 / 



22. (Vx + y/y)\ 23. (V* + ^~=J . 24. (| - ^f . 

26. (- * + y~ 2 Y. 27. (x» - y 2 )\ 28. (^ + ~^) 5 . 29. (» + 2/ + z) 2 . 

30. (* + 2y + z) 2 . 31. (a; 2 + x + l) 4 . 32. (a 2 - a - l) 3 . 

In each of the problems from 33 to 42, find the indicated term. 
33. (1 - x)*, 8th term. 34. (m + n)", 10th term. 

35. (a + 26) 12 , 5th term. 36. (a - &)» 3rd term. 

- — J , 4th term. 38. {x - y) 10 , term involving t/ 4 . 

^ J , term involving — • 

y *" * y 

41. (y/x - y/y) 12 , middle term. 42. (2x - #) 7 , middle terms. 



5 



Logarithms 



5-1. DEFINITION OF A LOGARITHM 

We shall assume here that the laws of exponents stated in 
Chapter 4 for rational exponents ark valid also for irrational 
exponents. The definition of a base raised to an irrational power is 
beyond the scope of this book. However, let us make the assump- 
tion that, if b and x are real numbers, with b positive, a corre- 
sponding number designated by b x exists. Without giving an 
explicit rule for computing b x , let us assume that all laws of expo- 
nents established in Chapter ~4 are valid generally for real powers. 
Finally, let us assume that, corresponding to any two positive real 
numbers & and n, where b ¥=■ 1, there exists a unique real number x, 
such that n = b x . We can then give the following definition. 

Definition. If n = b x , where 6 is a positive real number different 
from 1, then x is called the logarithm of n to the base 6. We write 
x = log & n. The following table shows both forms of several equiva- 
lent statements. 



Exponential 
Form 


2 3 =8 . 


41/2 = 2 


6 "81 


5° = 1 


Logarithmic 
Form 


log 2 8 = 3 


log 4 2 = \ 


log 3 gf = - 4 


log 5 1=0 



We shall restrict n to positive numbers, since negative numbers 
do not have real logarithms. ' 

For any positive base b, we have 6° = 1 and 6 1 = 6. Hence, it fol- 
lows from the definition of a logarithm that 

log b 1 = and log& b = f l. 

Another valuable, relationship results from combining the two 
equations n = b x and x = log& n. Replacing x in the first equation by 
its value from the second equation, we have 

n = &log„ % m 

For example, 2 10 ** 8 = 8, and 10 lo *i<> w = x. 

101 



1 02 Logarithms Sec. 5-1 

Example 5-1. Find n, if logs n = 2. 

• Solution: Write the given equation in exponential form, as follows: 

„ w = 32 - 

Hence, n = 9. 

Example 5-2. Find the base 6, if log* 4 = 2/3. 

Solution: In exponential form, the given equation is b 213 =4. Raise both sides 
to the 3/2 power and recall that b > 0. Then 

(£2/3)3/2 = ^ = 43/2 # 

Therefore, 6=8. 

Example 5-3. Find x f if logi/ 8 32 = x. 

Solution: Writing the equation in exponential form, we have 

Express 1/8 and 32 as powers of 2, and get 1/8 = 1/2 3 = 2" 3 , and 32 = 2 5 . Hence, 

(2-3)* = 2 5 , or 2- 8 * = 2 5 . 
From this, we have - 3z = 5, and a; = - 5/3. Therefore, logi/g 32 = - 5/3. 



5-2. LAWS OF LOGARITHMS 

Since a logarithm is an exponent with respect to a given base, 
the rules for operating with logarithms are the same as the laws 
of exponents. These laws, expressed in terms of logarithms, have 
the following form. 

Law I. The logarithm of a product equals the sum of the loga- 
rithms of its factors. The logarithmic form is 
(5-1) log& (m • n) = log 6 m + log6 n. 

Proof: To prove this equation, let 

x = log& m and y = log& n. 
Then 

m = b x and n = b y . 
Multiplying, we have 

mn =6^. 
Hence, . 

log 6 (mn) = x + y = log 6 m + log& n. 

Law II. The logarithm of a quotient equals the logarithm of the 
dividend minus the logarithm of the divisor. The logarithmic 
form is 



(6-2) log 6 (~) = log 6 m - log 6 n. 



Sec. 5-2 Logarithms 1 03 

Proof: The proof follows : Let 

x = logb m and y = log& n. 

Then ,„ , ,., 

m = b* and n = b y . 

Dividing, we have _ 

- = b<~y. 
n 

Hence, /m . 

iogft ^— y = a - 2/ = log b m ~ l°g& n * 

Law III. The logarithm of a power of a number equals the expo- 
nent times the logarithm of the number ; that is, 
(5-3) log*, (n fc ) = k log 6 n. 

Proof: The first step in the proof is to let 

x = logb n. 

Then . 

n = o z . 

Raise both sides to the fcth power and obtain 

This relationship, when written in logarithmic form, becomes 

log 6 (n k ) = to. 

Replacing # by its value, we have 

log& (n k ) = & log6 n. 

The student should note carefully the difference between log& (n k ) 
and (log&n) fc . 

Law IV. The logarithm of a root of a number equals the loga- 
rithm of the number divided by the index of the root ; that is, 

(5-4) log 6 y/n = t log b n. 

Proof: This equation follows as a corollary of law III. By the 
definition of a fractional exponent, we have ^/n = n 1/k . Hence, by 
law III, 

log 6 y/n = log& (n llk ) = ^log* n. 

f a/51 
Example 5-4. Express log2 — ^- as a linear combination of logarithms. 

iSerftiton: log 2 ^~P = log 2 VSl - log 2 3 4 = log 2 (3 • 17V'» - log 2 3 4 
= log 2 3 1 '? + log 2 17 1 ' 2 - log 2 3 4 . 



104 Logarithms Sec. 5-2 

Example 5-5. Express 2 logio 3 - x 1°8 10 x + 1°S 10 2/ as a single logarithm. 

Solution: 2 logio 3 - ~ ^°Sio a; + logio y = logio 3 2 - logio x 112 + logio 2/ 

= logi (3 2 -y) - logio x 1 ' 2 



, 3 2 2/ , 9y 



Example 5-6. Transform the equation logo # + y = log a sin x into an equation 
free of logarithms. 

Solution: By transposing, we get 

sin x 
2/ = loga sin * - logo x = logo -— • 

Change to the following exponential form: 

sin x 
w = • 



EXERCISE 5-1 

In each of the problems from 1 to 12, write the equation in logarithmic form. 
1. 2 3 = 8. 2. 2« = 64. 3. 3 4 = 81. 4. 10° = 1. 

5. 10 3 = 1000. 6. 10- 3 = 0.001. 7. 256 1 ' 8 = 2. 8. 216 1 ' 3 = 6. 

9. 100 - 6 = 10. 10. y=ep. 11. 10 v = x. 12. 10 lo « v = x . 

In each of the problems from 13 to 21, write the equation in exponential form. 

13. logs 64 = 2. 14. logs 125 = 3. 15. log 2 ^ = - 6. 

16. log* ^ = - 4. 17. log 7 343 = 3. 18. log 9 729 = 3. 

two 

19. logio 10,000 = 4. 20. logio 0.0001 = - 4. 21. log 4 8 = 3/2. 

In each of the problems from 22 to 33, find the indicated value of x. 
22. logo 3 = x. 23. log 2 64 = x. 24. log 4 x = 0. 

25. log* 4=2. 26. logo.s x = - 1. 27. log 3 x = 1. 

28. log, 81 = 4. 29. log, 100 = - 2. 30. log, ^ = 5. 

7 
31. logo 243 = x. 32. log 64 x = - ^ • 33. log, a = 2. 

In each of the problems from 34 to 39, use the laws of logarithms to write the 
expression as a single logarithm. 

34. log& 2-3 log* 5 + log 6 7. 35. log& 4 + log* w - log* 3+3 log* r. 

19 1 2^ 

36. ± log* 7 + 1 log* 4 + i log* 3. 37. - 5 log* 23 + 12 log* y • 



38. 3 log* 2 + log* 13-2 log* 5. 



Sec. 5-4 Logarithms 105 

39. ^ log* {u - Vu 2 - a 2 ) - ^ logb (u + s/u 2 - a 2 ) + log& a. 

40. Find the logarithm to the base b of the area of a circle in terms of the logarithms 
of 7r and the radius. 

41. The time T for a pendulum of length I to make one oscillation is T = ir *y - > 

where g is a constant representing the acceleration due to gravity, a) Find 
logb T in terms of the logarithms of tt, I, and g. b) Find log& I in terms of the 
logarithms of w, 1\ and g. 

42. The area of a triangle with sides of length a, b y and c is given by the formula 



K = V$(-s — a) (s - 6) (s — c), where s is the semi-perimeter ^ (a -f 6 + c). 
Find log& X in terms of the logarithms of combinations of a, 6, and c. 

43. The positive geometric mean G of n positive numbers Xi, x 2 , ■ • • . x n is defined 

by the relationship 

, n log 6 Ji + log 6 x 2 + • • • -f log& z n 
log* G = -s 



Show that G = v^xi.^ • • • x„. 

5-3. SYSTEMS OF LOGARITHMS 

As we mentioned in Section 5-1, any positive number b different 
from 1 may be used as a base in a system of logarithms. However, 
only two bases are widely used in practice. 

The common, or Briggs, system of logarithms, named for Henry 
Briggs (1556-1631), employs the base 10 and is used for ordinary 
computations. 

The natural, or Napierian, system of logarithms, named for 
John Napier (1550-1617), is generally used in calculus and theo- 
retical work, and employs the more convenient irrational base 
e = 2.71828 

In this book, when the base is not indicated, it is understood to 
be 10. Thus, log n means logi n, and the word logarithm will mean 
common logarithm unless otherwise stated. 

5-4. COMMON LOGARITHMS 

In Table 5-1, we begin with a list of powers of 10, give equivalent 
logarithmic forms, and from these determine the form of the loga- 
rithm of a number that is not an exact power of 10. It should be 
mentioned that the logarithm is an increasing function; that is, 
as n increases, log n increases. Another way of stating the condi- 
tions is to say that if a> b then log a > log 6. 



106 



Logarithms 

Table 5-1 



Sec. 5-4 



Exponential form 


Logarithmic form 


Logarithm of the number 


10 3 = 1000 


log 1000 = 


3.000 


<-log354. = 


2 + decimal 


10 2 = 100 


log 100 = 


2.000 


<-log 35.4 = 


1 + decimal 


10 1 = 10 


log 10 = 


1.000 


<-log3.54 


+ decimal 


10° = 1 


logl 


0.000 


<-log 0.354 = 


— 1 + decimal 


io- 1 = 0.1 


log 0.1 = 


- 1.000 


<-Iog 0.0354 = 


— 2 + decimal 


io- 2 = 0.01 


logO.Ol = 


- 2.000 


<-log 0.00354 = 


— 3 + decimal 


io- 3 = 0.001 


log 0.001 = 


- 3.000 







From Table 5-1 it can be seen that the following statements are 
true: 

The logarithm of an integral power of 10 is an integer. 

The logarithm of a number which is not an integral power of 10 
consists of two terms or parts : an integral part, called the charac- 
teristic; and a positive or zero decimal part, called the mantissa, 
which is determined from a table of mantissas. 

Thus, since log 10 = 1 and log 100 = 2, we may expect the loga- 
rithm of any number between 10 and 100, that is, a number between 
10 1 and IO 2 , to be 1 plus a positive decimal part. For example, we 
shall find that the logarithm of 35.4, which number lies between 10 
and 100, is equal to 1.5490, to four decimal places. In this case, the 
characteristic is 1 and the mantissa is .5490. 



5-5. RULES FOR CHARACTERISTIC AND MANTISSA 

A study of Table 5-1 reveals that the characteristic changes as 
the position of the decimal point changes in the sequence of digits 
0035400. The first entry in the column headed "Logarithm of the 
number" is 

log 354 = 2. + decimal. 



Sec. 5-5 Logarithms 1 07 

In the number 354, or 354.0, the decimal point is two places to the 
right of the first non-zero digit, 3 (reading from left to right) ; 
the corresponding characteristic is 2. 
The second entry is 

log 35.4 = 1. + decimal. 

In this number, 35.4, the decimal point is one place to the right of 
the first non-zero digit (reading from left to right) ; the corre- 
sponding characteristic is 1. 

Similarly, we note that the zero characteristic corresponds to the 
position of the decimal point immediately following the first non- 
zero digit. This position of the decimal point is called the standard 
position. We may now formulate the following rule for 
characteristics : 

Rule for Characteristics. If the decimal point is in standard posi- 
tion, the characteristic is zero. For every other position of the 
decimal point, the characteristic is equal to the number of places 
the decimal point has been shifted from the standard position. The 
characteristic is positive if the shift is to the right, and is nega- 
tive if the shift is to the left. 

We shall now see that the mantissa remains the same for all 
numbers having the same sequence of digits. Let us again consider 
the sequence of digits 0035400. Any number containing this 
sequence can be written 3.54 • 10 n , where n is a positive or negative 
integer or zero and depends on the position of the decimal point. 
Suppose that we consider the form log 3.54 = 0.5490. Then the 
logarithm of any number containing this sequence is 

log (3.54 • 10 n ) = log 3.54 + log 10" 
= n + log 3.54 
= n + 0.5490. 

Thus, a shift of the decimal place in the number affects only the 
characteristic n, and the mantissa remains the same for the same 
sequence of digits. 

EXERCISE 5-2 

In each of the problems from 1 to 16, find the characteristic of the logarithm oh 
the given number. 
1. 34.63. 2. f 3.463. 3. 34630. 4. 268.1. 

5. 0.1340. 6. 2637. 7. 0.00346. 8. tan 42°8'. 

7 821 
9. sin 63°41'. 10. 0.000001. 11. 378364. 12. ]^^' 

13. cot 81°13'. 14. sin 84°53'. 15. cos 61°43'. * 16. sec 24 b 8'. 



1 08 Logarithms Sec. 5-5 

In each of the problems from 17 to 24, place the decimal point in the sequence of 
digits 7314 corresponding to the given characteristic. 

17. 3. 18. - 2. 19. 0. 20. 1. 

21. 6. 22. - 5. 23. - 3. 24.-1. 

5-6. HOW TO WRITE LOGARITHMS 

As stated in Section 5-4, the mantissa of a logarithm is always 
positive or zero, whereas the characteristic may be a positive or 
negative integer or zero. A positive characteristic or a zero char- 
acteristic can readily be combined with a given mantissa. For 
example, the logarithm of 354 is written 2.5490. But when the 
characteristic is negative, say — fc, where 1 ^ k ^ 10, it is more con- 
venient to write it in the form (10 — k) — 10. Let us consider the 
logarithm of 0.00354. The characteristic is —3, but the mantissa is 
regarded as positive. We could write log 0.00354 = — 3 + 0.5490. 
For convenience in computation, however, we write log 0.00354 in 
the form (10 - 3) + 0.5490 - 10 = 7.5490 - 10, or 17.5490 - 20, 
and so on. 

Note. It would be incorrect to write log 0.00354 = — 3.5490, for 
this notation means — 3 — 0.5490 and would imply that the mantissa 
is negative. To perform certain computations, it is convenient to 
write the logarithm 7.5490 - 10 in the form -2.4510, which equals 
— 2 — 0.4510. It is important to note that the decimal part of the 
number —2.4510 is not the mantissa of the logarithm of 0.00354, 
since it is not positive. 

5-7. HOW TO USE A TABLE OF MANTISSAS 

The following examples will illustrate the procedure in finding 
the logarithm of a number with the aid of a table of mantissas. The 
student should work through each example, determining the char- 
acteristic from the position of the decimal point in the number and 
determining the mantissa by referring to Table III at the end of 
this book. 

Example 5-7. Find log 46.7. 

Solution: The characteristic is -f 1. To find the mantissa, locate 46 in the 
column in the table headed N, and then go to the right to the column headed 7. 
Here we find the mantissa .6693. So the complete result is log 46.7 = 1.6693. 

Interpolation. If the number consists of more than three digits, 
the mantissa is found from Table III by means of interpolation. 
Since the method of interpolation is the same as that described in 
Section 3-10 for the table of trigonometric functions, there will be 
no further discussion of it here. 



Sec. 5-7 



Logarithms 



109 



Example 5-8. Find log 0.03426. 

Solution: The characteristic is — 2. The mantissa is found by interpolation, 

since the number 3426 has more than three digits. It lies — of the way between 

the mantissas of 3420 and 3430, as shown in the accompanying tabulation: 

Number Mantissa ■ 

f f 3420 .5340 1 1 

6 * 

10 1 3426 .5340 + x J ) 13 

3430 .5353 

Since the difference between the mantissas of the two numbers in the table is 13, 
we have a 

x=^ (13) = 7.8. 

This is rounded off to 8, and the amount to be added to 0.5340 is given by x = 8. 
Hence, the mantissa is .5348 and log 0.03426 = 8.5348 - 10. 

Finding Antilogarithms. The number which corresponds to a 
given logarithm is called the antilogarithrn. That is, if \ogn = x, 
then n is the antilogarithrn of x and is written antilog x. 



Example 5-9. Find n, if log n = 1.8710. 

Solution: Search through the body of Table III to locate the mantissa .8710. 
The corresponding number, from the columns headed N and 3, is 743. Since the 
characteristic is 1, n = 74.3. 



Example 5-10. Find antilog 7.5349-10. 

Solution: The mantissa .5349 is not in Table III but lies between .5340 and 
.5353. To these correspond, respectively, numbers whose digits are 3420 and 3430. 
We may indicate the work in tabular form as follows: 

Number Mantissa 



10 



3420 
3420 + x 



.5340 
.5349 



13 



3430 .5353 

From this, we see that * 

x_ _ 9_ 

10 ~ 13 # 
Therefore, x = 6.9, or 7 after rounding off. Hence, the sequence of digits in the 
desired number is 3427. Since the characteristic is - 3, the untilogarithm of 
7.5349-10 is 0.003427. 



110 



Logarithms 
EXERCISE 5-3 



Sec. 5-7 



In each of the problems from 1 to 30, find the common logarithm of the given 
number. 

3. 105. 4. 0.0843. 5. 0.00621. 

8. 12.3. 9. 0.354. 10. 0.0781. 

13. log 7.03. 14. log 95.5. 15. log 695. 

18. 0.003821. 19. 0.7777. 20. 7,437. 

23. 0.08788. 24. 15.46. 25. cos 16°13'. 

26. sin 10°18'. 27. tan 41°33'. 28. sec 64°16'. 29. cos 82°14'. 30. cob 31°16'. 

In each of the problems from 31 to 50, find the antilogarithm of the given number. 



1. 35. 


2. 98. 


6. 9.63. 


7. 23,100. 


11. 0.0663. 


12. 1,630. 


16. log 6.31. 


17. 0.007001 


21. 3.142. 


22. 1.414. 



31. 1.6665. 32. 4.4857. 33. 9.4183-10. 

35. 2.7024. 36. 7.7388-10. 37. 9.4409-20. 

39. 1.8401. 40. 3.9552-10. 41. 2.4658. 

43. 9.7367. 44. 4.996O-10. 45. 8.7863-10. 

47. 0.6584. 48. 3.0150. 49. 5.0300-10. 

Solve for x in each of the following equations: 
51. 10- = 4. 52. 10* = 2.019. 

54. 10 01 = x. 55. ^/10 = x. 

57. v"l0* = z. 58. 10 1 - 314 = x. 

60. 10-*' 2 = 0.0123. 61. 10 1 -* = 0.2346. 



34. 0.0645. 
38. 6.3404-10. 
42. 1.9501. 
46. 9.8821-20. 
50. 0.1504. 

53. 10 2 * = 7.132. 
56. 10^« = x. 
59. 10-* = 0.003146. 
62. 10 2 *- 3 = 0.6735. 



5-8. LOGARITHMIC COMPUTATION 

The fundamental laws of logarithms given in Section 5-2 are 
applied in the following examples to illustrate the application of 
logarithms to computation. 

Example 5-11. Find the product (0.0246) • (1360). 

Solution: Let x = (0.0246) • (1360). Then 

log x = log 0.0246 + log 1360. 
log 0.0246 = 8.3909 -10 
log 1360 = 3.1335 
logo; = 11.5244-10 
= 1.5244. 
Hence, by interpolation, we have x = 33.45. 



Sec. 5-8 Logarithms 1 1 1 

Example 5-12. Evaluate (0.506)- 1 ' 3 . 

Solution: Let x = (0.506) -»'» = (05 q 6)1/3 • Then 

log x = log 1 - log (0.506) 1/3 
= log 1 - (1/3) log 0.506 
= log 1 - (1/3) (29.7042-30) 
= log 1 - (9.9014-10). 
log 1 = 10.0000-1Q 
1/3 log 0.506 = 9.9014-1Q 
log x = 0.0986. 
Therefore, x = 1.255 by interpolation. 

Alternate Solution: Let z = (0.506) - 1 ' 3 . Then 

log x = - (1/3) log 0.506 

= - (1/3) (29.7042-30) 
= - (9.9014-10) 
= - ( - 0.0986) 
= 0.0986. 
Therefore, x = 1.255 by interpolation. 



n i r to ^ i x (0.352) (1.74)2 

Example 5-13. Evaluate =~^ — • 

^0.00526 

Solution: Let x denote the desired value. Then 

log x = log 0.352 + 2 log 1.74 - (1/3) log 0.00526. 

We find that log 0.352 = 9.5465-10, log 1.74 = 0.2405, and log 0.00526 = 7.7210-10. 

log 0.352 = 9.5465-10 (1/3) log 0.00526 = (1/3) (27.7210-30) 

2 log 1.74 = 0.4810 = 9.2403-10. 

log numerator = 10.0275-10 

log numerator = 10.0275-10 
log denominator = 9.2403-10 

log x = 0.7872. 
Interpolating, we have x = 6.126. 



253 y 14 



(253' 
174 

Solution: Let x = ( j=j j . Then 



log x = 1.14 [log 253 - log 174]. 
log 253 = 2.4031 
log 174 = 2.2405 

Then °' 1626 - 

log x = 1.14 (0.1626) = 0.1854. 
Therefore, x = 1.532. 



112 



Logarithms 



Sec. 5-8 



EXERCISE 5-4 

In each of the problems from 1 to 30, perform the indicated computation using 
logarithms. 



1. (3.142)(2.718). 
29.34 



3, 



683.5 



5. ^(5.678) 2 . 



7. \/(0.003468)\ 



9. 



a V (5,321, OOP)* 
V (36,250) 4 



11. (63.84)2(0.0134). 



13. V(168.3) (14.21). 



15. V(23,310) 2 -(20,180)2. 
(Hint: Factor the radicand.) 

t » /123.4V* 



19, 



21 



1 • 3 • 5 • 7 « 19 • 31 
2* 4- 8- 16 •32*64* 

V (8.013) (0.034) 

23. (18,120) > (1,0 ^° 4 "" 1 > 

25. </68l3 - (0.8123)- 3 ' 4 . 



27. ^log 0.08614. 

29. l(3.864)-3-i3 + (0.841)-o-wp'«. 



2. (13.25) (26.80). 
4. (0.8134) 1/3 . 

6. (16.83) 3 '*. 

42,860 ' 

10. (4.313)(3,068)(0.000642). 
12. (8.364)(321.5)+(- 42.63). 



14. V(213.6) 2 (43.98)2. 
1A /68.34y' 5 

Mara?; • 



18. 



20, 



3,6 42 y/ (21.36) 3 
(1,083) 4 (0.0813) 3 



1.63 V 0. 



483 



38, 
22. (1.08)1°. 



8103 



24. 3,648 (1.03) 36 . 
26. 0.083 ' 412 . 



28. log 16.84 - ^483.6. 



30. 



(83.14)" 4 - 3 - (0.8134)2/3 



(0.6841)1/2 



31. If e = 2.718, find log e, log \/e, log - , e* f and t. 



32. Find the geometric mean of 564.3, 8634, 0.1349, 8.316, and 42.61. (Hint: See 
Problem 43, Exercise 5-1.) 

33. Find the area of a circle of radius 6,381 feet. 

4 

34. The volume of a sphere is V = « rr z . Find the volume of a sphere of radius 
3621 feet. 6 

35. Find the radius of a sphere whose volume is 8423 cubic feet. 

36. Find the length of a pendulum which makes one oscillation in 1 second, if 
g = 980 centimeters/sec 2 . (Hint: See Problem 41, Exercise 5-1.) 



Sec. 5-9 Logarithms 1 1 3 

37. Find the area of a triangle with sides 6,384 feet, 5,680 feet, and 2,164 feet long. 
(Hint: See Problem 42, Exercise 5-1.) 

38. The stretch s of a wire of length I and radius r by a weight m is given by the 

relationship s = — %r > where g is the gravitational constant and k (Young's 

modulus) is a constant for a given material. Find how much a copper wire of 
length 120 centimeters and of radius 0.040 centimeters will be stretched by a 
weight of 6,346 grams, if g = 980 and k is 1.2 • 10 12 for copper wire. 

39. The current i flowing in a series circuit with a resistance of R ohms and L 
henrys t seconds after the source of electromotive force is short-circuited is 
given by the relationship i = Ie~ RtlL , where I is the current flowing in the 
circuit before the short circuit. If i = 10 amperes, R = 0.1 ohms, t = 0.25 
seconds, and L = 0.05 henrys, find /. (Take e = 2.718.) 

40. If n is a positive integer, n! has been defined as the product 1 • 2 n. 

When n is very large, it is difficult to compute this product. However, Stirling's 
formula gives (approximately) n! = n n e~ n -\/2irn. Use this formula to estimate 
9!, and compare the result with the true value which you should calculate 
exactly. Do the same with 30!. (Hint: log (n!) = n log n — n log e + 

2 log 2 + ~ log 7T + - log n). 

41. If the rate of depreciation r per year is constant, the scrap value S after n years 
of a machine with first cost C is given by the formula S = C(l — r) ft . Find the 
scrap value after 10 years of a machine which originally cost $10,000, if 20 
per cent per year is written off as depreciation. 



5-9. CHANGE OF BASE 

It is sometimes desirable to change from one logarithmic base to 
another. Suppose there is available a table of logarithms to some 
known base b (say 10, for exampte), and we wish to find the loga- 
rithm of a number n to some other base a. We then let x = log* n ; 
whence, by definition, n = 6*. Similarly, if we let y = log n, then 
we have n = a y . 

It follows that a y = b*, and our problem reduces to solving this 
equation for y. Taking the logarithm of both sides to base 6, we 
have 

t y log* a = x log& b. 
But log6 6 = 1. Therefore, 

(5-5, "»-Eji- 



1 1 4 Logarithms Sec. 5-9 

Example 5-15. Find log* 125, where e = 2.7183, by using a table to the base 10. 
. Solution: By (5-5), 

loge 125 = -p 

6 logio e 

It is usually easier to multiply than to divide. Since division by logio e is a fairly 
frequent operation in practical work, it should be noted that ^aoao = 2.3026, 

and the result can be obtained by multiplying by 2.3026 instead of dividing by 
0.4343. Thus, 

log. 125 = (2.0969) (2.3026) = 4.828. 



EXERCISE 5-5 

Find each of the following logarithms by using a table of common logarithms: 
L log« 10. 2. log 9 100. 3. log2 e. 4. log e tt. 

5. log* e. 6. log* 10. 7. log 2 64. 8. log 20 1000. 

9. log 20 100. 10. logioo 64. 11. log e 8. 12. logo.i 50. 

13. logi25 1000. 14. loge 20. 15. logo.02 0.04. 16. logiooo 100. 



6 Right Triangles and Vectors 



6-1. ROUNDING OFF NUMBERS 

Numbers that arise in the applications of trigonometry are 
usually not exact, but are sufficiently accurate for a given purpose. 
Numbers of this kind are called approximate numbers, and the 
degree of accuracy of such a number is indicated by how many 
significant figures it contains. Reading from left to right, the 
significant figures in a number are the digits starting with the first 
non-zero digit and ending with the last non-zero digit, unless it is 
definitely specified that the zeros on the right are significant. Thus, 
in the numbers 2.405, 0.002405, and 240500, the digits 2, 4, 0, and 5 
are significant figures. The zeros after the 5 in 240500 may or may 
not be significant figures. 

When it is desired to indicate whether final zeros are significant 
or not, scientific notation is often used. Thus, in 2.405 • 10 5 , the 
last significant figure is 5 ; in 2.40500 • 10 5 , the final two zeros are 
regarded as significant. 

To round off a number in which the last desired significant figure 
is in the units place or in any decimal place, drop all digits that lie 
to the right of the last significant figure. It is sometimes necessary 
also to increase the last digit in the retained part by 1. 

If the first digit in the dropped part is less than 5, the last digit 
in the retained part is left unchanged. If the first digit in the 
dropped part is greater than 5 or if that digit is 5 and it is followed 
by digits other than 0, the last digit in the retained part is 
increased by 1. Whten the dropped part consists of the digit 5 
alone or the digit 5 followed only by one or more zeros, we shall 
use the following procedure as an arbitrary rule in this book: If 
the last digit retained is odd, this digit is increased by 1; if it is 
even, it is left unchanged. This rule, although popular, is inferior 

115 



1 1 6 Right Triangles and Vectors Sec. 6—1 

to common-sense rules in many cases. For example, if .245, .165, 
.485, and .725 are to be rounded off to two decimal places and then 
added, it would be more sensible to round off two of the numbers 
in one direction and two in the other direction. 

To round off a number in which the last significant figure will lie 
to the left of the units place, first drop all digits to the right of the 
place occupied by the last significant figure, and replace each 
dropped digit to the left of the decimal point by a zero. Also, 
either leave the last digit of the retained part unchanged or 
increase that digit by one, in accordance with the directions just 
given for dropping only a decimal part. For example, if 2533.62 is 
to be rounded off to three significant figures, the result is 2530 ; and 
if 487,569 is to be rounded off to three significant figures, the result 
is 488,000. 

There are two rules that are generally adopted by computers in 
working with approximate numbers in order to guard against 
retaining figures that may indicate a false degree of accuracy : 

1. In adding or subtracting approximate numbers, round off the 
answer in the first place at the right in which any one of the 
given numbers ends. 

2. In multiplying or dividing approximate numbers, round off 
the answer to the fewest significant figures found in any of 
the given numbers. The numbers entering a problem involv- 
ing multiplication or division may be rounded off before the 
computation is begun. If these numbers are rounded off, 
they should have one more significant figure than the answer 
is to have. 

While these rules point in the right direction, it should be men- 
tioned that rounding off computed quantities to as many significant 
figures as there are in the given numbers does not necessarily 
produce the degree of accuracy implied by the results. The subject 
of accuracy of computation with approximate numbers is somewhat 
complicated and beyond the scope of a book at this level. 

6-2. TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES 

One of the simplest, yet important, applications of trigonometry 
is in the solution of right triangles. A right triangle has, in addi- 
tion to the 90° angle, five other parts. These are two acute angles 
and three sides. If we know the length of any two sides, or either 
acute angle and any one side, the triangle can be solved; that is, 
the unknown parts can be found. 



Sec. 6-3 



Right Triangles and Vectors 



117 



hV 






•fl 






c s 




a 


A 




b 


C " 



To solve problems involving parts of tri- 
angles, we shall find it helpful to be able to 
express the trigonometric functions of an 
acute angle A of a right triangle ABC in 
terms of the sides of that right triangle. To 
derive suitable relationships, let us place the 
acute angle in standard position, as shown in 
Fig. 6-1. 

In the right triangle ABC the side AC, which is the abscissa of 
the point B, becomes the side adjacent to the angle A ; the side CB, 
which is the ordinate of B, becomes the side opposite to angle A ; 
and the side AB, which is the radius vector to B, becomes the 
hypotenuse. If we let the lengths of the side adjacent, the side 
opposite, and the hypotenuse be represented by the symbols b, a, 
and c, respectively, we may express the six functions of the acute 
angle A in terms of a, b, and c as follows : 

. A __ side opposite __ a 

Sin Jx — "-; ~ — 

hypotenuse c 

— ^de adjacent _ 6 

hypotenuse "~ c 

side opposite __ a 



(6-1) 



(0-2) 



cos A 



(6-3) 
(6-4) 



(6-5; 



(6-6) 



tan A = 



esc A 



sec A 



cot A = 



side adjacent 
hypotenuse 

side opposite 
hypotenuse 

side adjacent 

side adjacent 



side opposite a 

By using (6-1) to (6-6), we can express the trigonometric func- 
tions of an angle of a right triangle without reference to any 
coordinate system, since the ratios of the sides remain the same 
regardless of the position of the triangle. 



6-3. PROCEDURES FOR SOLVING RIGHT TRIANGLES 

When solving a right triangle in which two parts are known, it 
is advisable to arrange the work systematically and to follow £ 
definite procedure consisting of the following steps : 

1. Draw a figure reasonably close to scale, and indicate the 
known parts. 

2. Write an expression containing a trigonometric function 
which involves the two known parts and one unknown part. 




1 1 8 Right Triangles and Vectors Sec. 6—3 

3. Find the selected unknown part from this equation. 

4. Find all other unknown parts of the triangle by a similar 
procedure. 

5. Check all results. 

Whenever possible, select a trigonometric function that gives a 
solution by means of a multiplication rather than a division. 

In the following illustrative examples, 
the acute angles are represented by the 

letters A and B, and the right angle is ^^ a -658 

denoted by C, while the small letters a, 
b, and c, respectively, represent the sides 
opposite them. 

Example 6-1. Solve the triangle ABC, if A = 28°20' and a = 658. 

Solution: The triangle is drawn approximately to scale in Fig. 6-2. The unknown 
parts are the angle B and the sides b and c. 
Since A + B = 90°, we have B = 90° - 28°20' = 61°40'. 
To find the side b, we may apply either (6-3) or (6-6), since both equations 

involve the unknown b and the known parts A and a. We shall use cot A = - 

because it enables us to proceed to the solution by means of a multiplication rather 
than a division. Since A = 28°20' and a = 658, we have 

4 = cot28 ° 20 ' 

Then 

b = 658 cot 28°20 / 
= (658) (1.855) = 1220.59. 
Xn this example, we take b equal to 1221. This result is rounded off to four digits. 
To find the side c, we shall use (6-4), or esc A = - • We have, therefore, 

esc 28°20' = 7^3 • 
658 

Hence, 

c = 658 esc 28°20 / 

= (658) (2.107) = 1386. 

1221 
To check, we may use the relation cos A = -r^^ — 0.8802. Hence, A = 28°20'. 

Checking by means of the Pythagorean theorem yields the result b 2 = c 2 - a 2 
= (c - a) (c+a) = (728) (2044) = 1488232, whereas b 2 = (1221) 2 = 1490841. 
These values of b 2 agree when they are rounded off to three significant figures. 

Note. In most situations where we must solve triangles, we are 
dealing with measured quantities, which are necessarily approxi- 
mate. Therefore, our answers can be no more accurate than the 




Sec. 6-3 Right Triangles and Vectors 1 19 

data we begin with. If the original data are 
approximate, our answers must be rounded off to 
the degree of accuracy indicated by the data. 

In example 6-1, for instance, the answers may 
be given as b = 1220 and c = 1390, both rounded 
off to three significant figures. 

Fig. 6-3. 

Example 6-2. Solve the triangle ABC, if b = 250 and c = 371. 

Solution: The conditions are shown in Fig. 6-3. Since - = cos A, we have cos A 

2^0 c 

= =£ = 0.6738. Therefore, A = 47°38' and B = 90° - 47°38' = 42°22'. 

To find a, we have a choice of combining the unknown a with either b or c; 

hence, we may use either -r = tan A or - = sin A . We shall illustrate, in order, 

the computation with each of these equations, thus providing a check on our work. 
Using (6-3), we have 

^o = tan47 ° 38 - 

Hence, a = 250 tan 47°38' = (250) (1.096) = 274.0, or 274 when rounded off to 
three figures. 
Using (6-1), we have 

~L = sin 47 o 38 / b 

Hence, a = 371 sin 47°38' = (371) (0.7388) = 274.09 or 274 when rounded off to 
three figures. 

EXERCISE 6-1 

In each of the problems from 1 to 16, solve the right triangle. 

1. a = 12, A = 33°. 2. b = 168, A = 38°16'. 

3. b = 62.4, B = 71°10'. 4. a = 42, c = 76. 

5. a = 3.187, b = 6.249. 6. 6 = 63.21, B = 83°36\ 

7. a = 4.318, B = 67°16'. 8. b = 827.6, c = 963.4. 

9. a = 9.863, A = 36 21'. 10. 6 = 16.32, 5 = 87°10\ 

11. 6 = 78.21, A = 43°17'. 12. a = 43.21, c = 63.75. 

13. a = 123.6, 6 = 783.1. 14. a = 36.83, A = 57°44'. 

15. 6 = 2.312, B = 40°57'. 16. a = 389.3, 6 = 62.34. ? 

17. A wire stretches frorrj a point on level ground to the top of a vertical pole. It 
touches the ground at a point 15 feet from the foot of the pole and makes an 
angle of 63° with the horizontal. Find the height of the pole and the length 
of the wire. 

18. A ladder 40 feet long rests against a vertical wall. If its footjs 5 feet from the 
base of the wall, what angle does it make with the ground? 



120 



Right Triangles and Vectors 



Sec. 6-3 



19. A ladder 65 feet long is placed so that it will reach a window 35 feet above the 
ground on one side of a street. If the foot of the ladder is held in the same 
position and the top is moved to the other side of the street, it will reach a 
window 28 feet above the ground. How wide is the street from building to 
building? 

•20. The grade of a hill is the tangent of the angle the hill makes with the horizontal. 
Find the grade of a hill which is 275 feet long and which rises 120 feet. 

21. To find the width of a river, a surveyor sights on a line across the river between 
two points A and B on opposite banks of the river. He then runs a line AC 
perpendicular to AB. He finds that AC is 250 feet and angle ACB is 42°17'. 
How wide is the river? 

22. Find the length of a side of a regular hexagon and the radius of the inscribed 
circle, if the radius of the circumscribed circle is 10 feet. 

23. An airplane rises 560 feet while flying upward for 2,387 feet along an inclined 
straight-line path. What is the angle of climb? 

24. A pendulum 4.5 inches in length swings through an arc of 28°. How high does 
the bob rise above its lowest position? 

25. A man 6 feet tall is walking along a straight horizontal path directly away from 
a lamp post 10.5 feet high. How far is he from the post at a certain instant 
when his shadow is 5 feet long? 




Horizontal 



Fig. 6-4. 



6-4. ANGLES OF ELEVATION AND DEPRESSION 

Let the line AB in Fig. 6-4 be a level or horizontal line, and let 
an observer at the point A see an object at the point C. If the 
object C is above the horizontal line AB, then the angle BAC meas- 
ured up from the horizontal to the line of sight AC is called the 
angle of elevation to C from A. If the object C is below the hori- 
zontal line AB, then the angle BAC measured down from the* 
horizontal to the line of sight AC is called the angle of depression 
to C from A. 



Example 6-3. From a point on the ground 300 feet from the base of a building, 
the angle of elevation to its top is 22° 10'. How high is the building? 

Solution: In Fig. 6-5, we have b = 300 and A = 22°10'. By (6-3), ~g 

= tan 22°10'. Hence, a = 300 tan 22°10' = (300) (0.4074) = 122.22. So the build- 
ing is 122 feet high. 



Sec. 6-5 



Right Triangles and Vectors 



**E 



121 





*»e 



Fig. 6-5, 



6-5. BEARING IN NAVIGATION AND SURVEYING 

In marine and air navigation and in surveying, the direction in 
which an object is seen is expressed by the bearing or azimuth of 
the line of sight from the observer. The bearing of a line is the 
acute angle which its direction makes with a meridian or north- 
south line. Such angles are sometimes called quadrant angles, or 
quadrant bearings. To describe the bearing of a given direction, 
we first write the letter N or S, then the acute angle, and finally 
the letter E or W. The letters depend on the quadrant in which 
the given direction falls. Thus, the bearings of the lines OA, OB, 
and OC in Fig. 6-6 are N 60° E, S 37° W, and N 45° W, respectively. 

The azimuth of a line differs from its bearing only in that the 
azimuth is the angle measured from 0° at north in a clockwise 
direction. An azimuth may have any value between 0° and 360°. 
Thus, in Fig. 6-6 the azimuths of the lines OA, OB, and OC are 
60°, 217°, and 315°, respectively, measured clockwise from the 
north. This method of measuring directions is coming into more 
frequent use than that of quadrant bearings. We note also that the 
term bearing is often used instead of azimuth. Thus, we may speak 
of the bearing of an object regardless of whether we mean azimuth 
or bearing as here defined. 



Example 6-4. A ship heads due east from a dock at a speed of 18 miles per hour. 
After traveling 30 miles it turns due south and continues at the same speed. Find 
its distance and bearing from the dock after 4 hours. * 

Solution: In Fig. 6-7, let A be the point at which the dock is located, let B be the 
point where the ship turns south, and let C be the position of the ship after 4 hours. 

30 5 

Since the number of hours required to travel from A to B is ^ = - > the number 

5 7 7 

of hours spent in travel from B to C is 4 - = = ^ • Hence, a =" 18 • ^ = 42. 



122 


Right Triangles and Vectc 


From the figure, 


42 
tan = r^ and sec = 


Therefore, 


= 54°28', 


and 





Sec. 6-5 



30 



b = 30 sec 54°28' = (30) (1.721) = 51.6. 
Hence, the distance from the dock is 52 miles and the bearing of the line AC is 
(44°28' or S 35°32' E. 

6-6. PROJECTIONS 

Often it is desirable to consider direction along a line segment. 
Thus, if Pi and P 2 are the end points of a segment, we shall under- 
stand PiP 2 to mean the directed segment from P x to P 2 , the direc- 
tion being specified by the order in which the end points are named. 
The non-negative length of the segment PiP 2 is denoted by |PiP 2 |. 

Frequently a directed segment PiP 2 may lie on a line, such as a 
coordinate axis, on which a positive direction has been specified. 
Then the positive direction on the line may agree with the direction 
from Pi to P 2 , or the two directions may be opposite to each other. 
The directed length of the segment PiP 2 is equal to |P X P 2 | when the 
directions agree or when Pi and P 2 coincide and is equal to -|PiP 2 | 
when they disagree. Since the context will make the meaning clear, 
we shall designate the directed length of the segment PiP 2 also 
by PiP 2 . 




IF 



mr 2 ) 



C(o,y x ) 



J*2 ( x z*yz) 




/ A(x v Q) B(x 2> 0) 

Fig. 6-9. 



*»x 



We recall that the projection of a point on a given line is the foot 
of the perpendicular dropped from the point to the line. If A in 
Fig. 6-8 is the projection of Pi on the line I, and if B is the projec- 
tion of P 2 on I, then the directed segment from A to B is the 
projection on I of the directed segment P\P 2 . We draw P X M parallel 
to I, or perpendicular to P 2 B, to show the angle between I and 

PA. 



Sec. 6-6 



Right Triangles and Vectars 



123 



We shall now assume that a positive direction has been specified 
on the line I. Then, since PiM = AB, it follows immediately from 
trigonometry that 

AB = | PiP 2 I cos 0, 

where 6 is the acute angle between the positive end of I and the 
positive half -line determined by the directed segment P1-P2. In Fig. 
6-8 it is considered that I is positively directed toward the right. 
The result just given can be applied, as seen in Fig. 6-9, in find- 
ing the projections of P\P 2 upon the coordinate axes. The directed 
lengths of the projections upon the cc-axis and the 2/-axis are, 
respectively, 

(6-7) AB = I P1P2 I cos 0, 

(6-8) CD = I P1F2 I sin 0, 

where is the angle between OX and PiP 2 , as shown in Fig. 6-9. 
When the coordinates of the end points of the segment P\P 2 are 
known, the projections AB and CD are readily expressed in terms 
of these coordinates. From the definitions of horizontal and vertical 
distances given in Section 2-2, it follows that 

(6-9) AB = #2 — #i and CD = 2/2 — j/i. 



i 


iV 










C (0,2) ■ 




i 


P x (3,2) 




















#(7,0) 







U(3,0)'\ 






IX0.-5) ■ 








P 2 (7,- 












Fig. 6-10. 



Fig. 6-11. 



Example 6-5. What kre the projections of the segment P1P2 on the axes, if 
Pi = (3, 2) and P 2 = (7, - 5)? 

Solution: In Fig. 6-10, AB = z 2 - 21 = 7 - 3 = 4, and CD = t/ 2 - 2/1 
= -5-2=-7. Since AB is -f 4, we know that AB is directed to the right. 
Also, since CD = - 7, we know that CD is directed downward. 



1 24 Right Triangles and Vectors Sec. 6-6 

Example 6-6. A ladder 12 feet long leans against the side of a house and makes 
an angle of 67° with ground. Find its projections on the ground and on the side 
of the house. 

Solution: The conditions are represented in Fig. 6-11. Let I = 12 be the length 
of the ladder. The required projections are found as follows. 
The projection on the ground is given by 

x = I cos 6 = 12 cos 67° = 12 (0.3907) = 4.7. 
The projection on the side of the house is given by 

y = I sin = 12 sin 67° = 12 (0.9205) = 11.0. 



EXERCISE 6-2 

1. Two points A and B are 5,000 feet apart and at the same elevation. An airplane 
is 10,000 feet directly above point A. Find the angle of depression from a 
horizontal line through the airplane to point B and the airplane's distance 
from point B. 

2. The Washington monument is approximately 555 feet high. Find the angle of 
elevation to the top of the monument from a point that is 621 feet from the 
base of the monument and at the same elevation as the base. 

3. If a kite is 130 feet above the ground and 150 feet of string is out, find the angle 
of elevation to the kite, assuming the string to lie on a straight line, 

4. Find the angle of elevation to the sun if a flagpole 95 feet high casts a shadow 
63 feet long on horizontal ground. 

5. A boat leaves its dock and heads N 52° W for 4 hours at 14 knots (1 knot = 1 
nautical mile per hour = 6,080.4 feet per hour). It then turns and heads 
N 38° E for 3 hours at 16 knots. Find the boat's final bearing and distance 
from the dock. 

6. The grade of a certain railroad bod is 0.1095. How many feet does a locomotive 
rise while traveling 175 feet along the track? 

7. An approach must be built up to the end of a bridge which is 40 feet above 
ground. If the approach is to have a 10% grade and the original ground is 
assumed to be level, how far from the end of the bridge must the approach start? 

8. A surveyor wishes to find the distance between two points A and B separated 
by a lake. He finds a point C on the shore of the lake such that angle ACB is 
90°. He measures AC and BC and finds that AC is 640 feet and BC is 285 feet. 
How far apart are A and B? 

9. A smokestack is 175 feet from a building. From a window of the building the 
angle of elevation to the top of the stack is 28°10'. The angle of depression to 
its base from the same window is 24°30'. Assuming that the ground is level, 
find a) the height of the window above the ground and b) the height of the 
smokestack. 

10. Two ships leave the same port at the same time. One travels N 42° E at 25 
knots. The other travels S 48° E at 33 knots. How far apart are the two ships 
after 4 hours? 



Sec. 6-7 



Right Triangles and Vectors 



125 



6-7. SCALAR AND VECTOR QUANTITIES 

We shall at this point find it necessary to distinguish carefully 
between two kinds of quantities, namely scalar quantities and vector 
quantities. 

A scalar quantity is a quantity whose measure can be fully 
described by a number. It is a quantity which can be measured on 
a real number scale. For example, temperature is a scalar quan- 
tity, measured on the scale of a thermometer. Also we shall define 
the scalar components of the segment PiP 2 to be the projections on 
the coordinate axes, or the directed lengths x 2 — Xi and y 2 — Vi 
given by (6-9) in Section 6-6. The student should note, however, 
that the scalar components of P 2 P\ are not equal to those of P1P2. 
The components of P 2 P\ are x x — x 2 and y x — y 2 and are the nega- 
tives of the respective components of PiP 2 . 

A vector quantity, or simply a vector, is a quantity possessing 
both magnitude and direction. A vector may be represented by any 
one of a set of equal and parallel line segments. Algebraically, 
a vector is fully described by the scalar components of any segment 
representing it ; all such segments have the same components. We 
shall, in fact, call these the scalar components of the vector and shall 



£\ — i — i 



n 



\S. . . . 



IY 



i >*i — i i i i i 



O. 



Fig. 6-12. 



«D 



IF 



cr • 



i i i i i i i i i i i i — >-x 



-*x 



Fig. 6-13. 



enclose them in brackets. In Fig. 6-12, three representations of the 
vector [4, 3] are shown. The arrowhead indicates the order in 
which the end points of the line segment are named. 

We may denote a vector by a single letter in boldface type, say v, 
or may represent it geometrically by any one of the segments, such &s 
AB y OP, or CD. Here A, O, and C represent the initial points of the 
three segments, while the terminal points at the arrowheads are 
B, P, and D. Actually, any other segment with the same magnitude 
and direction could have been selected to represent the vector v. 
Instead of using a single letter in boldface type to denote a vector 
represented by a segment P\P 2 , we may also use the notation PiP* 



1 26 Right Triangles and Vectors Sec. 6-7 

Properties of Vectors. We have seen that a vector is unambigu- 
ously represented by any one of a set of equal and parallel line 
segments. Hence, any point may be taken as the initial point 
of a segment which represents a vector. If the origin is so chosen, 
the coordinates of the end^point P(x,y) are actually the scalar 
components of the vector OP. We can therefore give the following 
simple definition of the magnitude of a vector and its expression 
in terms of scalar components : 

Definition. The magnitude, or length, of a vector v is the length of 
any one of the segments representing v. 

Let v = [>i, v 2 "\ be the vector represented by OP in Fig. 6-13, 
where v x and v 2 are scalar components. Then we have |v| = \OP\. 
Since OP is the hypotenuse of a right triangle, 



(6-10) | v | = VV + *>2 2 . 

In case the scalar components are given by the coordinates (x, y) 
of the end point P of the segment, the magnitude of the vector is 
y/x*+y 2 . 

From the definition of a vector, it follows that two vectors are 
equal if and only if their respective scalar components are equal. 
For example, u = [u lt u 2 ] equals v = [v lt v 2 ] if and only if u x = v x 
and u 2 = v 2 . 

We also define a special vector = [0, 0] to be the zero vector. 
It corresponds to the exceptional case in which P 2 coincides with 
Pi and may be considered as represented geometrically by a seg- 
ment of length zero, that is, by a point. The zero vector may be 
regarded as having any direction whatsoever. 

If a vector v. = \v l9 v 2 "] is given, the vector — v = [— v u —^2] is 
defined to be the negative of v. Thus, if P X P 2 = v denotes a vector 
represented by the segment PiP 2 , then P 2 P X = -v denotes a vector 
having the same length as P\P 2 but oppositely directed, namely, 
from P 2 to Pi. We note that 

- (- y) = v. 

A unit vector is defined as a vector whose magnitude is unity. 

If u = \u l9 ^2] is any non-zero vector, then j-5- = L^L. , -^-1 

is a unit vector. 

Multiplication of a vector by a scalar is performed by multiplying 
the magnitude of the vector by the absolute value of the scalar, 
maintaining the direction of the vector if the scalar is non-negative 
and reversing it otherwise. Thus, by k[u u u^\ we mean the vector 



Sec. 6-7 Right Triangles and Vectors 1 27 

[ku lt ku 2 ]. Using the equation [m, U2] = k |~ > ^1 1 we can express 
any vector v = [i^, w 2 ] as proportional to a unit vector v/k if we 
choose k = V^i 2 + ^2 2 . Changing v to a unit vector v/fc is called 
normalizing v. 

Sums and Differences of Vectors. Of the many questions which 
arise in the study of vectors, the one of greatest importance for us 
at present concerns the addition of vectors. The sum, or resultant, 
of two vectors is defined to be the vector which has for its scalar 
components the sums of the scalar components of the two vectors. 
Thus, the sum of the vectors u and v is given by the relationship 

(6-11) U + V = [Wi + Vi, U 2 + V2I 

It is important to note that some of the laws of the algebra of 
numbers also hold for vectors. Thus, the commutative law is 

u + v = v + u. 

Also, the associative law of addition is 

(u + v) + w = u + (v + w). 

And, for every vector v, 

v + ( - v) = 0. 

That these laws are satisfied for vectors can be shown geometrically 
or can be seen from (6-11) by virtue of the known laws of addition 
of real numbers. 

In terms of components, the rule for multiplication by a scalar is 
given by 

fcv = [kviy kv2]. 

In consequence of this definition, the following algebraic laws are 
satisfied : 

k(vi + y) = fcu + k\; 

(k + m)u = fcu + ma) 
k(mu) = (fcm)u. 

To find the sum of two vectors geometrically, we proceed as 
indicated in Fig. 6-14. This graphical representation of the sum 
of two vectors by means of the triangle construction was probably 
suggested by the behavior of physical quantities represented by 
vectors, such as forces, displacements, velocities, and accelerations. 
Their addition is effected by the triangle law or the parallelogram 
law. 



128 




Right Triangles and Vectors 



Sec. 6-7 




*»X 



Fig. 6-14. 



Fig. 6-15. 



The following steps indicate the actual procedure. 

1. Select a segment representing one of the vectors, say u, with 
its initial point at the origin of the coordinate system. 

2. Place the initial point of a segment representing the second 
vector, say v, at the terminal point of the first segment. 

3. Draw the segment from the origin to the terminal point of the 
segment for v. This segment represents the resultant vector, 
u + v. 

The difference u - v of two vectors is defined in a manner 
analogous to that used in defining the difference of numbers. Thus, 

u — V = u + (— v). 

If u = [u u v^\ and v = [v u v 2 ~\ , then 

U — V = [Ui — Vi, U2 — V2]. 

To find the difference u — v of two vectors geometrically, we 
proceed as indicated in Fig, 6-15. The steps are as follows : 

1. Place segments representing u and v with their initial points 
at the origin. 

2. Place the initial point of a segment representing — v at the 
terminal point of the segment representing u. Note that the 
segment representing — v is parallel and equal in length to 
that for v, but has the opposite direction. 

3. Draw the segment from the origin to the terminal point of 
the segment for —v. This segment represents the vector 
u — v. 



Example 6-7. Find the sum and difference of the vectors u = [5, - 3] and 
▼ = [-2,1]. 

Solution: The required vectors are 

n + ▼ = [5 - 2, - 3 + 1] = [3, - 2], 
and 

u-v = |5-(-2), -3-l] = [7, -4]. 



Sec. 6-*7 



kight Triangles and Vectors 



129 



Example 6-8. Express u = [4, 3] as proportional to a unit vector. 
Solution: The expression representing the vector is 

U = [4,3]=5[|,|]. 

[4 3"! 
- t - . To check, 

T4 31. 
note that the magnitude of U > r is 

To add vectors analytically, we first find the scalar components 
of the vectors to be added. 
From Section 6-7, we know 
that if segments represent- 
ing u and v make angles a 
and (3, respectively, with 
the #-axis, the scalar com- 
ponents are given by the 
projections upon the coordi- - 
nate axes. Thus, as shown 
in Fig. 6-16, 

u x = | u | cos a, 
and 

v x = | v | cos j3 7 



fac,i>y,), 




Fig. 6-16. 



Uy = I U | Sin OL, 



vy = | v | sin j8. 

Then the components of the resultant will be given by the 
algebraic sums 

X = u z + v x , Y = Uv + vu. 

Thus, the magnitude of the resultant R is given by 



| R | = V^ 2 + Y 2 . 

Also, the direction angle 6 satisfies the relationship 

Y 

tan 6 = -y • 

To determine the quadrant of correctly, it is important to keep 
in mind and use the correct signs of X and Y. For example, tan 

= -^-i = 1 might lead one to an incorrect value 45° for $ instead of 

the correct third-quadrant angle 225°. 

The student should note that this (analytic) procedure and the 
geometric procedure (parallelogram law) give the same results. This 
can be shown by appropriately combining the scalar components of 
the vectors u, v, u + v, and u - v in Figs. 6-14 and 6-15, where we 
illustrated the geometric procedure. 



130 



Right Triangles and Vectors 



Sec. 6-7 




Fig. 6-17. 




*~E 



Example 6-9. A block weighing 500 pounds rests on a smooth plane making an 
angle of 25° with the horizontal, as indicated in Fig. 6-17. What force, parallel to 
the plane, is necessary to hold the block in position? 

Solution: Let the block be at A. The force due to the weight mayjbe represented 
by a vector acting vertically downward through A, such as vector AC. The com- 
ponent of AC which is parallel to the inclined plane and which must be overcome is 
represented by the projection AB of AC on the inclined plane. Since angle 
BAC = 90° - 25° = 65°, 

AB = 500 cos 65° = (500) (0.4226) = 211.3. 

Hence, a force of 211 pounds must be applied parallel to the inclined plane to keep 
the block from sliding. (It is assumed that three significant figures are appropriate.) 



Example 6-10. Find the magnitude and direction of the resultant of a force of 
110 pounds acting in the direction S 46°27' E and a force of 100 pounds acting in 
the direction N 29°14' W. 

Solution: The given forces are represented by vectors in Fig. 6-18. Let F denote 
the magnitude of the resultant force and 6 the angle it makes with OE. 

Since 90° - 46°27 / = 43°33' and 90° - 29°14' = 60°46', the components F x and 
F y of the resultant are found as follows: 

F. = 110 cos 43°33' - 100 cos 60°46' 
= (110) (0.7248) - (100) (0.4884) 
= 79.73 -48.84=30.89; 



F v = 



110 sin 43°33' + 100 sin 60°46' 
= - (110) (0.6890) + (100) (0.8726) 
= - 75.79 + 87.26 = 11.47. 



Now 



tan 6 



F x 



11.47 



= 0.3713, and esc 6 = 



F 



30.89 ~ "•"• *"' m ~ wv v ~" 11.47 
Therefore = 20°22', and F = 11.47 esc 20°22' = (11.47) (2.873) = 32.95. Hence, 
the magnitude of F is 33 pounds, and its direction is N 69°38 / E. 



Sec. 6-8 Right Triangles and Vectors 131 

EXERCISE 6-3 

L Draw a diagram showing three different segments representing each of the 

given vectors. Find the magnitude of each vector. 
a. [3, 4]. b. [12, 5], c. [-2, 4]. d. [- 3, - 5]. 

e. [1, - 1]. f. [5, 3]. g. [0, 4]. h. [0, - 1]. 

2. Express each of the vectors in Problem 1 in terms of a unit vector. 

3. In each of the following cases, add the given vectors. Find the magnitude and 
direction of the resultant. 

a. [1, 1] and [2, 3]. b. [4, - 2] and [1, - 10]. 

c. [2, 0] and [ - 6, 3]. d. [1, 2], [4, 3], and [0, 7]. 

4. In Problem 3, parts (a), (b), and (c), subtract the first vector from the second, 
and find the magnitude and direction of the resultant. 

5. In Problem 3(d), subtract twice the third vector from the sum of four times the 
first vector and twice the second vector, and find the magnitude and direction 
of the resultant. 

6. A force of 40 pounds acts at an angle of 63° with the horizontal. What are the 
vertical and horizontal components of the force? 

7. If a ship sails N 48° W at 30 knots, what are its westward and northward 
components? 

8. One force of 28 pounds acts vertically upward on a particle. Another force of 
43 pounds acts horizontally on the particle. What is the magnitude of the 
resultant force, and what is its direction? 

9. A barrel weighing 160 pounds rests on a smooth plane which makes an angle 
of 22° with the horizontal. Find the force parallel to the plane necessary to 
keep the barrel from rolling down the plane. 

10. Three forces act on a particle. One of 50 pounds makes an angle of 25° with the 
horizontal; a second of 60 pounds makes an angle of 50° with the horizontal; 
and the third of 75 pounds makes an angle of 230° with the horizontal. Find 
the magnitude and direction of the resultant force. 

11. Four forces act on a body. The forces are 30, 45, 50, and 65 pounds, and they 
make angles with the horizontal of 25°, 160°, 240°, and 330°, respectively. 
Assuming that all the forces lie in the same vertical plane, find the magnitude 
and direction of the force necessary to hold the body in equilibrium. The 
required force is equal in magnitude and opposite to the resultant of the given 
forces. 

12. A shell is fired at an angle of elevation of 37°. Its initial velocity is 2,500 feet 
per second. Find the horizontal and vertical components of its initial velocity. 

6-8. LOGARITHMS OP TRIGONOMETRIC FUNCTIONS 

So far in this chapter, we have considered the use of a table Qf 
natural trigonometric functions and have solved various problems 
involving right triangles. In many problems, however, the compu- 
tation is greatly facilitated by the use of logarithms to perform the 
numerical operations. For this purpose the values of the logarithms 
of the trigonometric functions are required. Table III might be 
used to obtain the logarithms of the functions found in Table II, 



132 



Right Triangles and Vectors 



Sec. 6-8 



but the work is considerably lessened by the use of Table IV at the 
end of this book, which gives the logarithms of the trigonometric 
functions at once. 

Table IV is a four-place table giving the logarithms of functions 
at intervals of 10 minutes from 0° to 90°. For the sine and cosine 
of any angle between 0° and 90°, the tangent of any angle between 
0° and 45°, and the cotangent of any angle between 45° and 90°, 
the value of the function is less than 1; hence, the logarithms of 
these functions are negative, and we must write —10 after the 
tabulated entry. For the sake of uniformity, 10 has been added 
to each of the other entries in the table. In using the table, 
therefore, 10 must be subtracted from every entry. 

The method of using Table IV is similar to that described for 
Table II and will be illustrated by the following examples. 



Example 6-11. Find log sin 23°10'. 

Solution: Since this angle is given in Table IV, we find that log sin 23° 10' 
= 9.5948-10. 

Example 6-12. Find log cot 51°27'. 

Solution: From Table IV we obtain the values for the following tabulation: 

f log cot 51°20' = 9.9032 - 10 1 1 

V \ \* 

10' { { log cot 51°27' = 9.9032 - 10 - x J > 26 

log cot 51°30' = 9.9006 - 10 



The tabular difference is 26. Since 51°27' is ~ of the way from 51°20' to 51°30', 
x = 0.7(26) = 18.2, and we have 

log cot 51°27' = (9.9032-10) - .0018 
= 9.9014-10. 



Example 6-13. Find the acute angle 6 if log tan = 9.7827-10. 

Solution: The positive part 9.7827 lies between the entries 9.7816 and 9.7845 
in Table IV. The procedure for finding 6 may be indicated as follows: 

log tan 31°10' = 9.7816 - 10 

11 



10' 



x 11 
Hence, j^ = ^9 , and 



log tan 31°(10 + x)' = 9.7827 - 10 
log tan 31°20' = 9.7845 - 10 

8 = 31°10' + 55UO') 
= 31°14'. ^ 



29 



Sec. 6-9 Right Triangles and Vectors 133 

EXERCISE 6-4 

In each of the problems from 1 to 15, find the value of the given logarithm. 

1. log sin 48°20'. 2. log sin 21°20'. 3. log cos 86°20'. 

4. log tan 88°30'. 5. log cot 10°20'. 6. log sec 43°50'. 

7. log sin 13°26'. 8, log sec 48°57'. 9. log tan 41°14'. 

10. log esc 78°32'. 11. log cot 68°43'. 12. log cos 18°18'. 

13. log esc 83°16'. 14. log cos 1°8'. 15. log tan 51°34'. 

In each of the problems from 16 to 35, find the angle (or angles) between 0° 

and 360°. 

16. log sin = 8.8059-10. 17. log tan = 8.3660-10, 

18. log cos = 9.9959-10. 19. log sin = 9.1697-10. 

20. log sec = 0.4625. 21. log cot = 0.4882. 

22. log tan = 9.8483-10. 23. log tan = 0.1430. 

24. log esc = 0.4081. 25. log sec = 0.3586. 

26. log sin = 9.9567-10. 27. log tan = 9.7648-10. 

28. log cos = 9.9755-10. 29. log cot = 9.8666-10. 

30. log sec = 0.1967. 31. log esc = 0.3370. 

32. log cos = 9.1860-10. 33. log cot = 1.5976. 

34. log sin = 9.9974-10. 35. log sec = 0.3870. 

6-9. LOGARITHMIC SOLUTION OF RIGHT TRIANGLES 

The solution of a right triangle by means of logarithms is exactly 
the same as by natural functions, except that 
for the actual numerical computation a table 
of logarithms of the natural functions is used 
in conjunction with a table of logarithms of 
numbers. The following example will illus- 
trate the procedure. 

Example 6-14. Solve the right triangle ABC, in which A = 21°40' and b = 8.43. 

Solution: The values of the known parts are indicated in Fig. 6-19. We see that 
B = 90° - 21°40' = 68°20'. To find the side a, we have 

tan21«40'=^, 

Hence, 

a = 8.43 tan 21°40', 
or 

log a = log 8.43 + log tan 21°40'. 
Arrange the work as follows: 

log 8.43 = 0.9258 (Table III) 

. log tan 21°40' = 9.5991-10 (Table IV) 

log a = 10.5249-10 
Therefore, a = 3.35. (Table III) 

To find side c, we have 

— = cos 21°40\ 
c 




1 34 Right Triangles and Vectors Sec. 6-9 

Hence, ^ 



cos 21°40' 
and 

log c = log 8.43 - log cos 21°40'. 

Arrange the work as follows: 

log 8.43 = 10.9258-10 (Table III) 

log cos 21°40' = 9.9682-10 (Table IV) 

log c = 0.9576 

Therefore, c = 9.07. (Table III) 

EXERCISE 6-5 

In each of the problems from 1 to 8, solve the given right triangle. 

I. b = 100, A = 31°. 2. c = 3.45, a = 1.76. 3. A = 25°20', a = 63.4. 
4. A = 88°17', c = 108.1. 5. c = 6.275, 5 = 18°45'. 6- a = 645.3, 6 = 396.3. 
7. J5 = 27°9', a = 36.13. 8. 6 = 98.34, B = 18°48'. 

9. If a railroad track rises 30 feet in a horizontal distance of one mile, find the 

angle of inclination of the track. 
10. A force of 341 pounds and another force of 427 pounds act at right angles to 
each other. Find the magnitude of the resultant force and the angle it makes 
with each of the forces. 

II. A force of 628 pounds acts at 180°, and a force of 237 pounds acts at 270°. Find 
the direction and magnitude of the resultant. 

12. An airplane is flying due east at a speed of 485 miles per hour, and the wind 
is blowing due south at 33.6 miles per hour. Find the direction and speed 
of the phane. 

13. The westward and northward components of the velocity of an airplane are 
363 and 487 miles, respectively. Find the direction and speed of the airplane. 

14. The eastward and southward components of the velocity of a ship are 10.4 and 
16.8 knots, respectively. Find the speed of the ship and the direction in which 
it is moving. 

15. A force of 2673 pounds is acting at an angle of 47° 13' with the horizontal. 
Find its horizontal and vertical components. 

16. A force of 162.4 pounds is just sufficient to keep a block at rest on a smooth 
inclined plane. If the block weighs* 783.1 pounds, find the angle at which the 
plane is inclined to the horizontal. 

17. Two tangents are drawn from a point P to a circle whose radius is 14.32 inches. 
If the angle between the tangents is 32°28', how long is each tangent segment? 

18. Two buildings of the same height are 11,640 feet apart. When an airplane is 
8,000 feet above one of them, what is the angle of depression to the other one? 

19. A cable which can withstand a pull of 10,000 pounds is used to pull loaded 
trucks up a ramp. If the angle of inclination of the ramp is 36°16', find the 
weight of the heaviest truck which can be safely pulled up the ramp with 
the cable. 

20. The angle of elevation from one point on level ground to the top of a flagpole 
is 45°28'. From a point on the ground 25 feet farther away the angle is 39°56'. 
How high is the pole? 



7 



Trigonometric Functions of 
Sums and Differences 



7-1. DERIVATION OF THE ADDITION FORMULAS 

Heretofore, we were concerned with relationships between trig- 
onometric functions of a single angle. We shall now establish 
certain fundamental identities involving two angles, in terms of 
the functions of the single angles. The following identities express 
functions of the sum and difference of two angles in terms* 6f the 
functions of the separate angles. 



(7-1) 
(7-2) 
(7-3) 
(7-4) 

(7-5) 
(7-6) 



sin (a + /3) = sin a cos /3 + cos a sin /3, 

cos (a + /3) = cos a cos /3 — sin a sin /3, 

sin (a — ]8) = sin a cos /3 — cos a sin j8, 

cos (a — 13) = cos a cos /3 + sin a sin /3, 
tan a + tan /3 



tan 



(« + ^) = 1 _ tanatan ^' 



, . tan oj — tan j3 
tan (a - j8) = — - 



tan a tan /3 

We shall now prove the formulas for the sine and cosine by using 
the derivation developed by E. J. McShane. 1 





+x 



Fig. 7-1. 



Fig. 7-2. 



1 E. J. McShane. "The Addition Formulas for the Sine and Cosine," 
American Mathematical- Monthly, Vol. 48 (1941), pp. 688-89. 

135 



136 



Trigonometric Functions of Sums & Differences 



See. 7-1 



iiK 




Ip [cos [<x - /3), sin (a- 0)] 




/ V* 




#\« \ 

) ^ * X 


O 


^HJp pa.o) 



Fig. 7-3. 



Let )3 and a be any two angles with the same initial side OW, as 
shown in Fig. 7-1. On their terminal sides we choose points P and 
Q, respectively, each at unit dis- 
tance from 0. 

Let d represent the distance 
from P to Q. We shall now make 
two computations for d 2 , using 
first OW, and then OP, as the 
#-axis. 

When OW is used as the #-axis 
of a coordinate system, as shown 
in Fig. 7-2, we find that the coordi- 
nates of P and Q are (cos /3, sin ft) 
and (cos a, sin a), respectively. 
Hence, by the distance formula, 

d 2 = (cos a — cos /3) 2 + (sin a — sin /3) 2 

= cos 2 a — 2 cos a cos /3 + cos 2 /3 + sin 2 a — 2 sin a sin (3 + sin 2 /3. 
Since cos 2 a + sin 2 a = cos 2 /3 + sin 2 /3 = 1 , 

d 2 = 2 — 2(cos a cos /3 + sin a sin /3). 

Let us now use OP as the a?-axis, as shown in Fig. 7-3. Then the 
coordinates of P are (1,0) and those of Q are [cos (a-/3), 
sin (a — /3) ] . Hence, 

d 2 = [cos (a - jg) - l] 2 + sin 2 (a - 0) 

= cos 2 (a - 0) - 2 cos (a - j8) + 1 + sin 2 (a - 0) 
= 2 - 2 cos (a - j8). 

Equating the two expressions for d 2 yields 

2 — 2(cos a cos + sin a sin j8) =2 — 2 cos (a — |8). 
Therefore, 
(7-4) cos (a — 0) = cos a cos + sin a sin 0. 

This establishes (7-4). 

Setting a = 90° in (7-4) , we find that 

(7-7) cos (90° - 0) = sin 0. 

If in (7-7) we let /3 = 90° - y, we have 
(7-8) sin (90° - 7) = cos 7. 

From (7-7), (7-8), and (7-4), we obtain 

sin (a + 0) = cos [90° - (a + 0)] 
= cos [(90° - <*) - 0] 
= cos (90° - a) cos + sin (90° - a) sin 0, 



Sec. 7-1 Trigonometric Functions of Sums & Differences 137 

or 

(7-1) sin (a + ft) = sin a cos ft + cos a sin ft. 

This establishes (7-1). 

Since cos (-ft) = cos ft and sin (-ft) = -sin ft, we have, as a con- 
sequence of (7-4), 

cos (a + ft) = cos [a - ( — j8)] 

= cos a cos (— ft) + sin a sin (— ft) 
or 

(7-2) cos (a + ft) = cos a cos ft — sin a sin j8. 

Similarly, from (7-1) it follows that 

(7-3) sin (a — ft) = sin a cos (— j8) + cos a sin ( — ft) 

= sin a cos ft — cos a sin ft. 

To prove (7-5), we use the relationship tan 6 = s and (7-1) 

COS 

and (7-2). We then obtain 

, , ns sin (a + ft) sin a cos + cos a sin 

tan (a + ft) = r~~7oT ~ 5 : : — 5 * 

cos (a + ft) cos a cos p — sin a sin p 

Dividing each term of the numerator and denominator by 

cos a cos ft, we have 

sin a cos cos a sin 

co s a cos cos a cos ft ___ tan a + tan ft ^ 

cos a cos ft sin a sin ft 1 — tan a tan ft 

xx cos a cos ft cos a cos ft 

Hence, , , . o 

,*, rN / , /on tan a + tan p 

(7-5) ton < a + fl = i- -ton«tanfl - 

We may obtain (7-6) in a similar manner from (7-3) and (7-4), 
or from (7-5). 

Example 7-L Find the exact value of sin 75°. 

Solution: Substituting 45° for a and 30° for in (7-1), we obtain 
sin 75° = sin (45° + 30°) 

= sin 45° cos 30° + cos 45° sin 30° 

_\/2 Vj \/2 l t 
~ 2 ' 2 + 2 V 
Therefore, ,^ 

sin 75° = ^ (\/3 + 1). 



138 Trigonometric Functions of Sums & Differences Sec. 7—1 

Example 7-2. Find the exact value of tan (a 4- P) if a is a second-quadrant 

3 5 

angle such that sin a = ■=■ , and P is a third-quadrant angle tan P = — • 

Solution: Let a be a second-quadrant angle for which y = 3 and r = 5. 

3 
Hence, x = — 4. It follows that tan a = - - • Using (7-5), we have 

_§+£ 

* / , a\ 4 1 2 16 

tan (a + 0) = T-gTy^T = ~ 53 ' 



-(-)(&) 



EXERCISE 7-1 

In each of the problems from 1 to 8, find the exact value of the given function. 
1. cos 75°. 2. tan 105°. 3. sin 135°. 4. sin 15°. 

5. tan 195°. 6. cos 195°. 7. tan 15°. 8. cos 105°. 

9. If a is a third-quadrant angle and p is a second-quadrant angle, and 

3 5 

sin a = - -= and cos P = -rr> find sin (a + 0), cos (a -f P), and tan (a + P). 

10. If tan a = ^ and a - 6 = 45°, find tan 0. 

11. If tan a = 3 and a + p = 180°, find tan p. 

3 1 

12. If tan a = j and tan = - > find sin (a + 0) and cos (a +0), where a and 

are first-quadrant angles. 

13. If cos a = - and cos P = *= > find sin (a — 0) and cos (a — 0), where a and 



are acute angles. 

If sin a = -= » tai 
o 

and cos (a -f- j8). 

„ . 3 

If sin a. = ■=• > 
5 

cos (a — p). 



4 5 

14. If sin a = ■= » tan = — — > and a and are both obtuse, find sin (a + P) 

t+Ph 

3 24 

15. If sin a = ■= > cos P = jr? > a is obtuse, and is acute, find sin (a — p) and 



2 1 

16. If cos a = - - > sin = - > and a and are obtuse, find sin (a + j8) and 

cos (a + P). 

3 5 

17. If sin a = - > cos = — j$ > a is obtuse, and P is in the third quadrant, find 

tan (a + P) and tan (a - p). 
Prove each of the following identities: 

a/2 

18. sin (a - 45°) = -5- (sin a - cos a). 19. cos (a - 7r) = - cos a. 



20. 



sin (a - 0) _ / tt\ 1 + tan a 

— : r— —• = cot — cot a. 21. tan I a + -t ) = ^ 7 * 

sin a sin j3 \ 4 / 1 — tan a 

22. cos 2 a = cos 2 a - sin 2 a = 2 cos 2 a - 1 = 1 - 2 sin 2 a. 



Sec. 7-2 Trigonometric Functions of Sums & Differences 1 39 

OO • O O • OA Sil1 2 a COS 2 Q! 

23. sin 2 a = 2 sin a cos a. 24. — = sec a. 



sin a cos a 



25. 



sin ( a + fl) + sin (a - g) __ sin (a + 0) _ , , . p 

— __ tan a< 26. ^ = tan a -f tan 0. 

cos (a + p) + cos (a - p) cos a cos 

27 s i n Cg + 0) _ tan g -f tan ft ^ ^ cos (a - g) __ 1 + tan a tan |8 ^ 

sin (a - P) tan a - tan cos Ja -f jS) ~~ 1 - tan a tan 

oq cos (g - g) _ 1 + tan a ta n cot a cot - 1 

29 ' ^(alj) " la^TTlanT ' "' COt ( " + ft = ^ot^T"c"oT^ " 

31. Bin (a 4-0) sin (a -0) = ^ ^ _ ^ 

cos 2 a cos 2 p 

32. sin (A + B + C) =sin A cos £ cos C + cos A sin B cos C 
-f cos A cos J3 sin C - sin A sin B sin C. 

33. sin 3 a = sin 5 a cos 2 a — cos 5 a sin 2 a. 

34. cos (a — 0) cos (a + 0) = cos 2 a — sin 2 = cos 2 /3 — sin 2 a. 

35. sin 2 ( a + t) "~ cos2 ( a + t) ~ sin 2 a> 

7-2. THE DOUBLE-ANGLE FORMULAS 

If we let fS ~ a in (7-1), (7-2), and (7-5), we obtain functions 
of twice a given angle in terms of the functions of the angle itself. 
Thus, we have the following identities : 
(7-9) sin 2 a = 2 sin a cos a, 

(7-10) cos 2 a = cos 2 a — sin 2 a, 

and 

t» t i \ o 2 tan a 

(7-11) tan 2 a = 



1 — tan 2 a 

We may obtain two other useful forms for cos 2 a from (7-10) 
by using in turn cos 2 a — 1 — sin 2 a and sin 2 a = 1 — cos 2 a. These 
forms are given by the identities 

(7-12) cos 2 a = 1 - 2 sin 2 a, 

and 

(7-13) cos 2 Q! = 2 cos 2 a - 1. 

The following illustrations give an indication of the possible 

applications of (7-9), (7-10), and (7-11). The student should 

study them carefully. 

sin 4 a = sin 2 (2 a) = 2 sin 2 a cos 2 a, 

/a\ . ot ot 

sin a = sin 2 I ~ ) = 2 sin ^ cos ^ > 

cos - = cos 2 1^1 = cos 2 - — sur ^ • 



1 40 Trigonometric Functions of Sums & Differences Sec. 7—2 

Example 7-3. Find the exact value of sin 120° by means of a double-angle formula. 

Solution: We use (7-9) to obtain 

sin 120° = sin 2(60°) = 2 sin 60° cos 60° 



=»(#G)-# 



Example 7-4. Derive a formula for cos 3 a in terms of cos a. 

Solution: Applying the identity for cos (a -f 0), and the double-angle formulas, 
we have 

cos 3 a = cos (a -f 2 a) = cos a cos 2a — sin a sin 2 a 
= cos a (2 cos 2 a - 1) - sin a (2 sin a cos a) 
= 2 cos 3 a — cos a — 2 sin 2 a cos a 
= 2 cos a (cos 2 a — sin 2 a) — cos a 
= 2 cos a (2 cos 2 a — 1) — cos a 
= 4 cos 3 a — 3 cos a. 

^ i « ^ t» ., . , ...sin 3 cos 3 n 

Example 7-5. Prove the identity — r-^ -r- — 2. 

r sin cos 

Solution: First combine the fractions on the left side and then reduce the result 
to the right side. Thus, 

sin 3 __ cos 3 __ sin 3 cos - cos 3 sin _ sin (3 0-0) 

sin cos ~~ sin cos "~ sin cos 

sin 2 2 sin 2 



sin cos sin 2 



= 2. 



7-3. THE HALF-ANGLE FORMULAS 

Functions of an angle in terms of the functions of twice that 
angle can be obtained directly from (7-12) and (7-13) . If cos 2 a = 
1 — 2 sin 2 a is solved for sin a, we obtain 



V 1 



cos 2 a 
sin a 



Also, by solving cos 2 a = 2 cos 2 a — 1 for cos a, we have 



;a = ±|A 



+ cos 2 a 
cos r " * " 



2 

Since these formulas may be equally well regarded as expressing 
functions of half an angle in terms of the functions of the given 
angle itself, the same relationship is retained if the identities are 
written 

(7-14) 
and 

(7-15) 



. a 
sin ^ = 


±l />- 


cos a 
2 


a 
cos 5 = 


±l /- 


■ cos a 
2 



Sec. 7-3 



Trigonometric Functions of Sums & Differences 



141 



These are the so-called half -angle formulas for the sine and cosine. 
From (7-14) and (7-15) we obtain, by division, 



tr, -,n\ a si n a/2 , A /T 

< 7 - 16 > tan 2 = ^572 = ± yr 



— cos a 



t/2 y 1 + cos a 

The algebraic signs in (7-14), (7-15), and (7-16) are deter- 
mined by the quadrant of a/2. 

If we rationalize, in turn, the numerator and the denominator of 
the right hand side of (7-16) , we obtain 



tan 



%=±n 



— POS 2 



cos* a 



(1 + cos a) 2 



v- 



(1 + cos a) 2 



or 

(7-17) 
Similarly 



a sm a 

tan ^ = j—. 

2 1 + cos a 



tan f = ± \/ { \ ~ ™f = ± |/E^! 
2 V 1 — cos 2 a f snr a 



or 

(7-18) 



tan 



a 



1 — cos a 
sin a 



The student will note that, in deriv- 
ing (7-17) and (7-18), we have 
dropped the ± sign in each formula. 
The validity of this step should be 
verified by consideration of the signs 
of tan a/2, sin a, and 1 ± cos a. Thus, 
tan a/2 and sin a necessarily have the 
same sign, while 1 =t cos a is non- 
negative. 

Example 7-6. Find the exact value of 
tan 22.5°. 




r-24 



Fig. 7-4. 



Solution: Since the exact values of the functions of 45° are known, we may use 
(7-16), (7-17) or (7-18). Selecting (7-18), we obtain 

* ™r Q j. 45° 1 - cos 45° 2 

tan 22.5° = tan — = — r-— — = 7T~ 

2 sm 45° y/2 



V2 



1 42 Trigonometric Functions of Sums & Differences Sec. 7—3 

24 
Example 7-7. Given tan 2a= — =- > where 2 a is a second-quadrant angle. 

Find sin a and cos a. 

24 

Solution: Since 2 a is an angle whose tangent is — =- » we may find a point on 

the terminal side of the angle with x = — 7 and y = 24, as shown in Fig. 7-4. Thus, 



r = V49 + 576 = 25. Therefore, 

a/1 +7 '25 4 



and 



COS Of = /l/ — 



-7/25 3 
2 5' 



a 
Example 7-8. Given that tan - = u, find sin a and cos a: in terms of w. 

Solution: Squaring both sides of (7-16), we have 

n a 1 — cos a 

tan 2 - = — • 

2 1 + cos a 

ol 1 " — cos a. 

Substituting u for tan - > we get u 2 = ■= — ■ • If we solve this equation for 

* \j 4.w 2 1 4- cos a 

cos a, we rind that 

1 - u 2 
cos a = — — - • 
1 -f u 2 

If we substitute this value of cos a in the relationship sin 2 a + cos 2 a = 1, 

we obtain 



i/] 7l - fc 2 \ 2 2?< 



Note that we can drop the ± sign, just as we did in (7-17) and (7-18), since 
tan a/2 and sin a always have the same sign. 



EXERCISE 7-2 

In each of the problems from 1 to 8, find the exact functional value by using an 
appropriate double-angle or half-angle formula. 

I. sin 22.5°. 2. cos 15°. 3. sin 120°. 4. cos 90°. 

5. sin 67.5°. 6. cos 67.5°. 7. tan 67.5°. 8. tan 60°. 

12 

9. It is known that cos 6 — — ^ and 6 is positive in the second quadrant. Find : 

lo 

a. sin 20. b. cos 6/2. c. tan 26. d. cot 26. 

e. sin 6/2. f. cos 26. g. sin 36. h. tan 40. 

40 

10. It is known that sin 6 = — — and 6 is positive in the third quadrant. Find : 

a. sin 26. b. cos 6/2. c. tan 26. d. cot 26. 

e. sin 6/2. f. cos 26. g. sin 30. h. tan 40. 

II. It is known that tan = - 3 and is positive in the fourth quadrant. Find: 
a. sin 20. b. cos 0/2. c. tan 20. d. cot 20. 
e. sin 0/2. f. cos 20. g. sin 30. h. tan 40. 



Sec. 7-4 Trigonometric Functions of Sums & Differences 1 43 

In each of the problems from 12 to 28, write the given expression in terms of a 
single function of a multiple of 0. Make use of appropriate formulas to reduce the 
answer to as few terms as possible. 

12. 2 sin ^ cos | • 13. cos* ~ - sin* j . 14. 2 cos* 30 - 1. 



2 tan 30 1C , .50 t _ . nik A /l - cos 40 
1 - tan* 30 ' 16 ' * " Sin2 T ' 17 - Sin 2$ V 2 



— *V. m ___. „ .._. /, 9n sin 20 COS 20 

ZU. : jr — n * 

sm cos 



18. (sin - - cos 2) • 19- cos 4 - sin 4 0. 

21 2 00 2cot0 ^ 1 - tan* 0/2 

cot - tan 1 + cot* 1 + tan* 0/2 ' 

04 4. Q , * ^ ok sin 0/ 2 o* 2 tan 0/2 

24. cot x + tan 5 • 25. - '-^r • 26. t— — — -~-rr • 

2 2 1 - cos 0/2 1 4- tan* 0/2 

27. j 7 C0S H • 28. cos* (0*) - sin* (0*). 

1 + cos 30 

Prove each of the following identities: 

29. sin 40-4 cos 0(sin 0-2 sin 3 0). 30. cos 40 = 8 cos 4 0-8 cos* + 1. 

31. sin 20 = (1 + cos 20) tan 0. 32. 1 + sin 20 = (sin + cos 0) 2 . 

00 1 - tan* 0/2 2(1 - cos 0) ***-*•* ■ a « <>* 

33. £-*— = . — — - • 34. 4 sin cos* = sin + sin 30. 

cos sin* 



7-4. PRODUCTS OF TWO FUNCTIONS EXPRESSED AS SUMS, AND SUMS 
EXPRESSED AS PRODUCTS 

By adding and subtracting corresponding members of (7-1), 
(7-2), (7-3), and (7-4), we obtain 

(7-19) sin (a + 0) + sin (a - j8) = 2 sin a cos 0, 

(7-20) sin (a + |8) - sin (a - 0) = 2 cos a sin 0, 

(7-21) cos (a + 0) + cos (a - 0) = 2 cos a cos 0, 

(7-22) cos (a + 0) - cos (a - 0) = - 2 sin a sin 0. 

If we reverse these identities, they become the following product 
formulas, which express given products of sines and cosines as 
sums or differences : 

(7-23) sin a cos = ^ [sin (a + 0) + sin (a - 0)], 

(7-24) cos a sin = 2 [sin (a + 0) - sin (a - 0)], 

(7-25) cos a cos = 2 [cos (a + 0) + cos (a - 0)], 

(7-26) sin a sin = - | [cos (a + 0) - cos (a - 0)]. 



1 44 Trigonometric Functions of Sums & Differences Sec. 7-4 

To obtain the sum formulas, which express given sums or differ- 
ences of sines and cosines as products, we first let 

a + /3 = x and a — (3 = y. 

Then, solving for a and /3, we have 

% + y j q x — y 

a = — p^- and j8 = — ^ • 

Substituting these values of a and /3 in (7-19), (7-20), (7-21), 
and (7-22), we obtain the sum formulas. These are 

(7-27) sin x + sin y = 2 sin (^-y^) cos (^-y^) > 

(7-28) sin a: - sin 2/ = 2 cos (^-y^) sin ( - ~ y J > 

(7-29) cos x + cos y = 2 cos (^y^) cos (^-y^) > 

(7-30) cos x - cos y = - 2 sin (^jpQ sin (^y^) • 

Example 7-9. Express sin 3 a cos 5 a as a sum of sines. 
Solution: Using (7-24) and replacing a by 5 a and by 3 a, we obtain 
cos 5a sin 3a = r[sin (5a + 3a) - sin (5a - 3a)] = -[sin 8a - sin 2a]. 

Example 7-10. Express cos 40 + cos 20 as a product of cosines. 
Solution: Using (7-29) and replacing x by 40 and y by 20, we have 
cos 40 -f cos 20 = 2 cos — - — cos — - — = 2 cos 30 cos 0. 

^i i ~ *i t^ .li • i x-x sin 7x — sin 5x , 
Example 7-11. Prove the identity = — — — — r- = tan x. 

COS (X ~j~ COS ox 



Solution: 



/ 7x + 5x \ . / 7x - 5x \ 

sin 7x - sin 5x _ V 2 / V 2 / 

(Ix + 5z\ (Ix - 5a:\ 
2 cos (— y— j cos (— y-J 

2 cos 6a; sin a: 



cos 7.t -I- cos 5a: n /7x + 5x\ (7x - bx 

n _ # _ i cog i 

- tan x. 



2 cos 6a; cos x 

EXERCISE 7-3 

In each of the problems from 1 to 10, write the given expression as a sum or 
difference of two sines or two cosines. 

1. sin 30 cos 40. 2. 2 sin 40 cos 20. 3. 2 sin 60 cos 40. 4. sin cos 40. 
5. cos 40 cos 20. 6. 2 sin 65° cos 15°. 7. sin 28° sin 20°. 8. cos 21° cos 31°. 
9. sin 50 sin 0. 10. sin 110 sin 30. 



Sec. 7-4 Trigonometric Functions of Sums & Differences 1 45 

In each of the problems from 11 to 20, write the given expression as a product of 
sines and cosines. Hint: In problems 18, 19, and 20, note that cos = sin (90° - 0). 



11. sin 30 + sin 20. 




12. cos — cos 40. 


13. sin 60 + sin 30. 




14. sin 40° + sin 20°. 


15. cos 80° - cos 20°. 




16. sin 30° - sin 80°. 


17. sin 40° + sin 25°. 




18. sin 64° + cos 38°. 


19. sin 40° + cos 44°. 




20. sin 65° - cos 33°. 


Prove each of the following identities: 






21. sin + cos = y/2 cos (d - -j) • 


22. 


x , x a cos (a — 0) 

tan a + cot = — r~^ • 

cos a gin p 


no cos0 , sin0 OQ 
23. 72 + Ta = cos 30. 

sec 40 esc 40 


24. 


sin , cos0 . ^ 
sec 30 esc 30 


25. Sin2 9 ! l + ^ta* 
cos 20 + cos 40 


26. 


sin 29 - sin A 
cos 30 + cos 


27>8 in«-f S in^ =tan a+0. 
cos a + cos p 2 







sin (# + t) + sin (* ~ t) = ^ Sin < 
29. sin (0 + -| ) - sin (0 - ~) = \/3 cos 0. 

cos(-| +0) - cos(-| - 0) = - V3sin0. 



30. 

, sin - cos + 1 , 

31. . 2 « ITTT = tan 7r • 

sm + cos + 1 2 

32. sin (a + j8) sin (a - j8) = sin 2 a - sin 2 j8 = cos 2 fi - cos 2 a. 

33. cos (a + fl) cos (a - jS) = cos 2 a - sin 2 j8 = cos 2 - sin 2 a. 



Graphs of Trigonometric 

O Functions; Inverse 

Functions and Their Graphs 




(-1,0) 



8-1. VARIATION OF THE TRIGONOMETRIC FUNCTIONS 

In Section 3-2, the trigonometric functions were defined in terms 

of the coordinates (x, y) of the 
point P(t), where the number t 
represents the directed length of 
the arc of a unit circle measured 
from the point (1,0). Later, in 
Section 3-9, an equivalent defini- 
tion was given in terms of an 
angle 9 in standard position. We 
shall now consider the variation 
of the trigonometric functions in 
the four quadrants as t, and with 
it 0, increases from to 2n. From 
Fig. 8-1, we can read off the vari- 
ations shown in Table 8-1. 
For example, by noticing the changes in y as t increases con- 
tinuously from to 27T, we find that sin t varies from to 1 in the 
first quadrant, from 1 to in the second, from to — 1 in the third, 
and from -1 to in the fourth. Similar considerations lead to the 
results for the cosine and tangent. 

Recalling that esc t = — - > we know that if either of these func- 

sin I 

tions increases the other decreases. Hence, the variation in esc t 
can be determined from the variation in sin t. Similarly, we may 
learn about the variation of sec t from that of cos t, and about the 
variation of cot t from that of tan t. 

We found in Example 3-2 that tan n/2 is undefined, which means 
that tan t has no value when t = 7r/2. For the sake of easier tabu- 

146 



Fig. 8-1. 



See. 8-2 Trigonometric Functions; Inverse Functions 

TABLE 8-1 
Variation of Trigonometric Functions 



147 





From 


To 


From 


To 


From 


To 


From 


To 


t 





tt/2 


w/2 


T 


T 


3ir/2 


3tt/2 


2ir 


sin t 





1 


1 








-1 


-1 





cos t 


1 








-1 


-1 








1 


tan t 





CO 


— 00 








00 


— 00 





CSC t 


00 


1 


1 


00 


— 00 


-1 


-1 


— CO 


sec t 


1 


CO 


— 00 


-1 


-1 


— 00 


00 


1 


cot t 


00 








— 00 


00 








— CO 



lation of this result in Table 8-1 we have employed the much 
used symbols oo (infinity) and -co. These symbols merely signify 
that in the neighborhood of tt/2 or one of its odd multiples the 
value of tan t is very large numerically. They are not to be used as 
numbers. 

8-2. THE GRAPH OF THE SINE FUNCTION 

To construct a graph representing the variation of the sine, we 
let x denote a real number or the value of an angle measured either 
in radians or in degrees, and we let y denote the corresponding 
value of the function. Corresponding values of x and y are plotted 




+x 



Fig. 8-2. 



148 



Trigonometric Functions; Inverse Functions 



Sec. 8-2 



as points on a rectangular coordinate system. We can infer the 
general appearance of the curve from the results summarized in 
the preceding section, but an exact representation is more readily 
obtained by using a table of sines. 

Let us now construct the graph of y = sin x from x = -tt/2 to 
x = 2n. The following values, found by the methods of Section 3-2, 
are used to obtain the curve in Fig. 8-2. 



X 


7T 

" 2 


T 

3 


7T 

" 6 





"6 


7T 


7T 

2~ 


2tt 
3 


5?r 
6 


y 


-1 


-0.87 


-0.5 





0.5 


0.87 


1 


0.87 


0.5 



X 


T 


7tt 
6 


4w 


3tt 
2 


5tt 
3 


llTT 

6 


2tt 


y 





-0.5 


-0.87 


-1 


-0.87 


-0.5 






While the choice of a scale is arbitrary, a better propor- 
tioned graph results if the same unit of length is used on both 
axes. The unit so selected will represent the number 1 on the y-axis 
and one radian on the #-axis. In terms of this unit a suitable length 
can then be marked off on the #-axis to represent 2tt or 360°. 

8-3. THE GRAPHS OF THE COSINE AND TANGENT FUNCTIONS 

Using the table of trigonometric functions, the student should 
make a table of corresponding values for y = cos x and one for 




*»x 



y - cos x 



Fig. S-3. 



Sec. 8-4 Trigonometric Functions; Inverse Functions 



149 




7T/2 



y - tan x 



37T/2 



27T 



+»x 



Fig. 8-4. 



y = tan x, similar to that used for y = sin x in Section 8-2. Study 
the graphs in Fig. 8-3 and Fig. 8-4 on the basis of the tables you 
have made. 

If we compare Fig. 8-3 with Fig. 8-2, we see that the graph of 
y = cos x may be obtained from the graph of y = sin x by moving 
the graph of y = sin x to the left a distance of tt/2 units. This fact 
can be checked by using the relationship cos x = sin (x + ir/2) 9 
from which it follows that cos = sin 7r/2, cos tt/6 = sin 27r/3, and 
so on. 



8-4. PERIODICITY, AMPLITUDE, AND PHASE 

The trigonometric functions are among the simplest of a large 
class of functions which are periodic. As a preliminary to defining 
periodic functions, we shall call attention to some examples of 
phenomena which recur periodically, such as the rotation of the 
earth about its axis, sound and water waves, the vibration of a 
spring, and many other vibratory and wavelike phenomena. The 
behavior of the object involved in a phenomenon of periodic nature 
determines the type of function that is required to represent it 
properly. We note particularly that, because of the recurrence 
characteristic of such a phenomenon, the values which the function 
assumes in any interval of given length are also taken on in any 
other interval of the same length. This statement apparently indi- 
cates that a function of x is periodic with period p if, for every 
value of x, the function returns to the same value when x is 
increased by p. More specifically, we state the following. 



1 50 Trigonometric Functions; Inverse Functions Sec. 8-4 

Definition. A function f(x) is said to be periodic if there is a non- 
zero number p for which 

f(x + p) =f(x) 

for all numbers x in the domain of f(x). Any such number p is 
called a period; the smallest positive number p satisfying the 
requirement is called the period. 

Evidently, if p is the period of f(x), then np is a period, for 
every integer n. 

Periodicity of Trigonometric Functions. No matter which of the 
two viewpoints is considered in the definition of the trigonometric 
functions, we shall see that, if g t ^ 2n f then 

any trigonometric function of (t + 2tt) = same function of t. 

According to the definitions given in Sections 3-1 and 3-2, P(t) 
and P(t + 2Tr) represent the same point on the circumference of 
the unit circle. To locate P(t) we start at (1,0) and proceed 
around the circle in the proper direction a distance of |£| units. To 
locate P(t + 2ir) we continue another 2tt units from the point P(t) . 
This merely adds another complete revolution, and we arrive at the 
same point P(t). 

If we consider the definitions of the functions in terms of angles, 
as given in Section 3-9, we note that the angle + 2n is coterminal 
with and that any trigonometric function has the same value for 
coterminal angles. 

Period of Sine, Cosine, Cosecant, and Secant. From a study of the 
sine curve, it is apparent that sin x assumes all values between —1 
and +1 as x, starting from any value, varies through 2tt units. In 
other words, the graph of the function repeats itself during each 
interval of length 2tt, for positive and negative values of x. Or 
stated more concisely, 

sin (x + 27r) = sin x. 

This is equivalent to saying that sin x is a periodic function of x, 
and that 2tt is a period. It remains now to show that 2tt or 360° is 
the smallest positive number p for which sin (x + p) = sin x and 
for which cos (x + p) = cos x. 

Since, by definition of a period p, sin (x + p) = sin x for any value 
x, we shall select for the purpose of our proof the particular value 
x = it/2. We then have 

. /7T .• \ . 7T 

sin (^r + v) = S1 *i y = 1- 
But since sin ( -=- + pj = cos p } it follows that cos p = 1. 



Sec. 8-4 



Trigonometric Functions; Inverse Functions 



151 



Hence, p must be an even multiple of tt. The smallest even multiple 
of 7r is 2tt, which must also be the smallest positive period of the 
sine function. 

In a similar manner, we find that 2n is also the period of the 
cosine function. Because of the reciprocal relationships existing 
between the sine and cosecant and between the cosine and secant, 
2tt is also the period of the cosecant and the secant. 

Period of Tangent and Cotangent, To find the period of the 
tangent, we write 

tan (x + V) = tan x } 

and we let x = 0. Then tan p = 0, and we find that tt is the period 
of the tangent. A similar argument shows that tt is also the period 
of the cotangent. 

Period of sin bx. We have just seen that the period of sin x is 
2tt. We shall now determine the period of sin bx, where 6 is a 
positive constant. That is, we want to know the smallest positive 
change in x which will produce a change of 2ir in bx. If p repre- 
sents this change in x, we can find p from the relationship 

b(x + p) = bx + 27T. 

Solving for p, we immediately find that 

2tt 
P = T - 

Thus, the period of sin bx is equal to the period of sin x divided by 
b, that is, 27r/&. 

Similarly, it can be shown that 2tt /b is also the period of cos bx, 
esc bx, and sec bx. It is also true that the period of tan bx and 
cot bx is n/b. 



2> 


J 










1 






\ / \ y=*sit\2x 











7T/2\ 


A 37r/2\ 


Att 


■■"^ A. 


-1 




1 \ 


y~*2 sinx \ / 






-2- 













Fig. 8-5. 



152 



Trigonometric Functions; Inverse Functions 



Sec. 8-4 



Let us consider the graph ofy = sin 2x shown in Fig. 8-5. Since 
the period of sin 2x is 2ir/b = 27r/2 = 7r, the function will assume 
the same range of values in the interval from to tt that sin x takes 
in the interval from to 2tt. 

Note that the graph of y = 2 sin x is similar in form to that of 
y = sin x, which is shown in Fig. 8-2. The period is 2tt for both 
curves. However, for any given value of x, the corresponding value 
of y in y = 2 sin x is twice as large as is the corresponding value of 
y in y = sin x for the same value of x. In the graph ofy — a sin x, 
where a > 0, the greatest value of y is a, and the smallest value of y is 
—a. The constant a is called the amplitude of the function or of the 
graph. Thus, the amplitude of the graph of y = sin x is 1, while that 
of the graph of y = 2 sin x is 2. 

In general, for the function y = a sin x, where a is a real number, 
the amplitude is equal to |a| and the period is equal to 2tt. Also, in 
general, for the graph ofy — a sin bx> where a and b are real, the 
amplitude is \a\ and the period is 2rr/b. 

Phase Angle. Since the graph of y — cos x may be obtained from 
the graph ofy = sin x by shifting it to the left a distance equal to 
7r/2 units, we say that the graph of y = cos x differs in phase by 
7r/2 from the graph ofy — sin x. The amount of horizontal dis- 
placement of two congruent graphs, amounting to tt/2 radians in 
this case, is called the phase difference, or the phase angle. The 
amplitude and the period are the same for y = cos x as for y = sin #. 




*-x 



/ » cos 2x - sin (2 .r + 7T/2) = sin 2 (.r + 7T/4) 

Fig. 8-6. 



Now consider the graph of y = cos 2# = sin 2(x + ~) in Fig. 

8-6. This curve may evidently be obtained from that ofy = sin 2x 
in Fig. 8-5 by a shift of 7r/4 units to the left in the x direction. 
Hence, the graph of y = sin (2x + ~) differs in phase from that 
of y = sin 2x by 7r/4 radians. 



Sec. 8-4 Trigonometric Functions; Inverse Functions 1 53 

A simple method for finding the phase displacement is to locate 
the point near the origin for which the function sin (2x+^) 
equals zero ; that is, to find the smallest numerical value of x that 

7T TV 

makes the quantity x + - zero. We have then sin2(#-f-) =0 

7T 4 4 

when x + - = or when x = — tt/4. Hence, the phase difference is 

7r/4 radians, and the shift of the graph is toward the left since the 
sign of x is negative. 

It can be shown that, in general, the phase displacement of any 
trigonometric function of (bx + c), where b > 0, is to the right or 
left by \c/b\ radians (or degrees). The direction of the displace- 
ment depends on whether c/b is negative or positive. 

Finally, we arrive at the conclusion that for the graph of y = 
a sin (bx + c) the amplitude is |a| and the period is 27r/b. Also, its 
phase differs from that of y = sin x by c/b. 

Because of its usefulness in many applications, we shall illustrate 
by means of an example a procedure for reducing an expression 
of the form A sin + B cos # to the form a sin (0 + a) . 



Example 8-1. Reduce A sin -f B cos to the form a sin (0 + «)• 
Solution: Write 

A sin + B cos = VA 2 + B 2 (- _ sin + — = cos d) • 

\\/^ 2 + B 2 VA 2 +B 2 / 

A B 

The absolute values of the coefficients — — and — =— cannot be greater 

VA 2 + £2 VA 2 + B 2 

than 1, and the sum of their squares is 1. Hence, they may be taken as the cosine 
and sine, respectively, of some angle a. Therefore, the expression A sin 9 -f B cos $ 
becomes 



y/A 2 + B 2 (cos a sin 6 + sin a cos 6) = VA 2 + B 2 sin (0 + a), 
where A = a cos a and J5 = a sin a. 



Example 8-2. Express % = sin 20 - V3 cos 20 in the form y = a sin (60 + a), 
and sketch the graph. 



Solution: Since A = 1 and £ = - \/3, we have y/A 2 + B 2 =2. Therefore, a, 
sin 20 + V3 cos 20 = 2^ sin 20 - ^ cos 20^. 



If we identify this result with the expression \JA 2 -f B 2 (cos a sin 20 + sin a cos 20), 

we have cos a = ^ and sin a = — - • 

angle and may be taken equal to - w/3. 



1 /*\ 

we have cos a = ^ and sin a = -~ • It follows that a is a fourth quadrant 



154 



Trigonometric Functions; Inverse Functions 



Sec. 8-4 



We have, finally, - _ i / *■ \ • 

y = i(sin 20 - V3 cos 20) = ± sin (20 - j J • 

The amplitude of this function is 5 > its period is 7r, and its phase angle is tt/6. 




-1/2^: 



j--l/2sin(20-7T/3) 



Fig. 8-7. 



Since the interval from (x/6, 0) to (7tt/6, 0) is one period in length, the part of the 
curve obtained for this interval may be repeated indefinitely in both directions to 
give the complete curve. The curve is shown in Fig. 8-7. (Here for convenience 
we have employed different units of length on the two axes.) 



EXERCISE 8-1 

In each of the problems from 1 to 24, find the period, amplitude, and phase angle 
of the trigonometric function. 

2. sin ~ • 

5. ^ sin j 6 • 

8. Braz- 
il. 3 sin 56. 
14. sec 96. 
17. tan (tt 6 + 4). 

19. esc (tt 6 + 7). 20. cot (2tt 6 - tt). 

22. 6 - cos 40. 23. 5 + 3 sin (26 - j\ • 



1. 


3 sin 0. 


4. 



C0S 2' 


7. 


4 tan 1 



10. 


sin 7r0. 


13. cot 60. 



16. 3 esc 



3x 



3. 


2 cos 0. 


6. 


2 cot 1 • 


9. 


COS -j-0 • 
4 


12. 


5 tan 7r0 


15. 


| cot 40. 



18. cos (30 - 2). 

21. 2 + sin 0. 

24. 4 + 2 cos (20 + -|) 



In each of the problems from 25 to 30, sketch the graph of the given function 
by constructing a table of values. 

25. y = cos •= x • 26. y = tan 5 z • 27. 2/ = 2 sin 3.r. 



28. 2/ = 5 cos 5 a: • 



6' 
29. y = 2 tan j x * 



30. 2/ = 3 sin 2z. 



Sec. 8—5 Trigonometric Functions; Inverse Functions 1 55 

Sketch each of the following graphs without constructing a table of values. 

2 1 

31. y = sin ^ x. 32. y = cos 4z. 33. y = tan ^ #. 

34. y = 2 cos 3.r. 35. y = 3 cos to. 36. */ = ^ tan ^ Xt 

11 14 / ir\ 

37. 2/ = o sin o Xm 38. y =z cos ^ Z. 39. y = 5 cos ^4x + y J 

40. ?/ = cos (x + 2). 41. y = sin L? ~ s) • 42. y = cos (3z - 2). 

43. y = sin (2x + 1). 44. ?/ = cos (27rx - ir). 45. 2/ = sin f j x + 7 J • 

46. ?/ = sin + cos 0. 47. y = sin — cos 0. 

- - 13 

48. y = V3 sin 20 + V5 cos 20. 49. y = 5 cos tt0 - - sin tt0. 

50. y = V2 cos 30-3 sin 30. 51. y = 2 sin - cos 20. 

8-5. INVERSE FUNCTIONS 

In Section 2-3, we defined a function by setting up a rule of 
correspondence between twa sets of numbers, X and Y, called the 
domain of definition of the function and the range set of the func- 
tion, respectively. The function was called single-valued if just 
one number y of the set Y is assigned to each number x of the set X. 
If more than one number of Y corresponds to some value of x, the 
function is multiple-valued. 

If we know that y = f(x), we may pose a reverse problem. We 
assume y to be given and ask for all corresponding values of x. 
Naturally, y is limited to lie in the range of the given function, 
since otherwise no x exists. The function which makes correspond 
to each such y all values of x for which y = f(x) is called the 
inverse function corresponding to the given function. 

We shall begin our discussion of inverse functions with an 
example in which X is the set of all real numbers, Y is the set of 
all non-negative real numbers, and the correspondence is deter- 
mined by the relationship 

y = x 2 . 

Ordinarily, we assign values to x in order to calculate values of 
x 2 . In this case, we have a rule of correspondence that assigns just 
one number y to ea$h chosen number x. Hence, y is a single-valued 
function of x. The s graph ofy = x 2 is shown in Fig. 8-8. 

Assume now that y is given and that we wish to determine cor- 
responding values of x. To do this we solve the given equation 
x 2 = y for x, and obtain two numbers x = y/y and* x = —y/y cor- 
responding to every. non-negative number y. Since the rule of 



156 



Trigonometric Functions; Inverse Functions 



Sec. 8-5 



correspondence assigns two values of x to each chosen number y, 
we see that # is a double-valued function of y. In this case, the 
admissible values of y are restricted to zero and the positive real 
numbers, while those of x comprise all real numbers, as was indi- 
cated in the specification for the sets X and Y. 




i 1 I i >*x 



H 1 1 1 1 1 1 — I >~X 



- V 2 



x~y 



Fig. 8-8. 



Fig. 8-9. 



In the study of mathematics, we generally prefer to use the sym- 
bol x to represent the independent variable and y to represent the 
dependent variable. To be consistent with this preference, we shall 
call y = x 2 and y = ±\/x inverse functions, each being called the 
inverse of the other. 

The graph of y = ±\/x is obtained by plotting that of x = y 2 , as 
shown in Fig. 8-9. We note that the roles of x and y are interchanged 
in the two equations y=x 2 and x = y 2 . Thus, we see that the 
curve in Fig. 8-9 is actually the curve of Fig. 8-8 with the axes 
interchanged and one of them reversed in direction. 

8-6. INVERSES OF THE TRIGONOMETRIC FUNCTIONS 

The Inverse Sine. Let us consider the function y = sin x and 
attempt to apply a discussion similar to that in Section 8-5. Here, 
as has been noted, the domain is the set of all real numbers, while 
the range is the interval -1 S j/ ^ 1. Referring to the graph of 
y = sin x in Fig. 8-2, and recalling the periodic properties of this 
graph, we see that, for every given number y such that — 1 :sS y ^ 1, 
there are infinitely many values of x such that y = sin x. To desig- 
nate the totality of all values of x such that y = sin x, we write 

-i 



x = sm~ 



y, 



which is read x is the inverse sine of y. The student should note care- 
fully that the symbol sin* 1 y must be distinguished from (sin y)-\ 

which equals - — or esc y. 
sin v 



Sec. 8-6 Trigonometric Functions; Inverse Functions 1 57 

Another notation that is frequently used to represent this inverse 
function is 

x = arc sin y, 

which is read x is the arc sine of y. 

In order to conform to the preferred practice of considering y 
as a function of x, we may designate the inverse of the sine func- 
tion by writing 

y = sin™ 1 x or y = arc sin x. 

We note the following properties of this inverse function. 

If # is a number such that \x\ > 1, then y does not exist. This 
property follows from the fact that the sine function takes on only 
the values from — 1 to 1. Hence, the inverse sine function sin -1 x is 
defined only when — 1 ^ x ^ 1. For example, sin -1 2 is not defined, 
since there is no number or angle whose sine is 2. 

If x is a number such that \x\ ^ 1, then y certainly exists. More- 
over, because of the periodicity of the sine function, there are 
infinitely many values of y = sin * x corresponding to every such 
value of x. For example, it x — 1/2, then y = sin -1 1/2 means that 
y is any number or angle such that sin y = 1/2. Then y may be 
taken as 7r/6, 57?/6, or any value that differs from these by integral 
multiples of 2tt. The totality of these values of sin -1 1/2 may be 
represented as 

— + 2mr and — + 2mr, where n — 0, ± 1, zb 2, =fc • • . 

We shall find it convenient to plot the graph of y — sin -1 x for a 
further study of the inverse function. Since y = sin 1 x and x = sin y 
express exactly the same relation between x and y, the graph of 
Fig. 8-2 may be used as a graph of the inverse function. We obtain 
the graph of y = sin - 1 x simply by interchanging the axes in Fig. 
8-2 and reversing one of them in direction. The result is shown in 
Fig. 8-10. 

The question now arises whether the y-axis can be subdivided 
into intervals within each of which y has just one value correspond- 
ing to each x such that \x\ ^ 1. One way of doing this is by select- 
ing a first interval 

Other intervals, such as -- ^ y g — and — ^— £ y £ — =- > are then 

selected. With the entire 2/-axis thus subdivided, we may think of 
the graph of y = sin 1 x as consisting of the graphs of infinitely 
many single-valued functions or branches. 



158 



Trigonometric Functions; Inverse Functions 

IF 



Sec. 8-6 




^•■iSin~*a? 



+»X 



y** cos"" 1 x 




*»X 



Fig. 8-10. 



Fig. 8-11. 



To avoid any ambiguity in later applications as a result of this 
multiple-valued property of the inverse sine, we shall often restrict 
y so as to make the function single- valued. There will then be but one 
value of y corresponding to each value of x such that sin y — x. We 
shall determine this value from the branch for which — -- ^ y ^ — « 

This branch is called the principal branch. The values of y chosen 
from this branch are called the principal values of the inverse-sine 
function and are represented by the equation 

y = Sin -1 x or y = Arc sin x. 

Note that in this case the initial letter of the name is capitalized. 

This restriction to the two quadrants containing the smallest 

numerical values of y results in a single-valued function y = Sin 1 x. 

The values of y are such that - ■— ^ y ^ ~ f or x in the interval 

-1 ^ x ^ 1. For example, we have Sin- 1 (-1) = -tt/2, Sin- 1 (-1/2) = 
-tt/6, Sin- 1 = 0, and Sin" 1 1 = tt/2. 

The Inverse Cosine. We shall next consider the function y = cos x. 
With the help of Fig. 8-3, we see that the range ofy = cos x is the 
interval — 1 ^ y ^ 1, and that for every y in this interval there are 
infinitely many values of x such that y = cos x. We are thus led to 
the inverse cosine function, which makes correspond to y all the 
values of x such that y = cos #. If again we interchange the sym- 
bols x and y, we may write the inverse cosine function as 



r-l 



y = cos^ 1 x or 



= arc cos x. 



Sec. 8-6 



Trigonometric Function^; Inverse Functions 



159 



This inverse function has the following properties. 

If # is a number such that \x\ > 1, then y does not exist, because 
there is no value of y for which |cos y\> 1. 

If x is a number such that \x\ ^ 1, then there are infinitely many- 
values of y designated by cos -1 x. 

The graph ofy = cos -1 x is shown in Fig. 8-11. The method for 
plotting it is similar to that used for graphing y = sin -1 x. We first 
write x — cos y. We then plot a cosine curve by proceeding as for 
Fig. 8-3, except that values of the independent variable y are laid 
off on the 7/-axis. 

The principal branch of the curve in Fig. 8-11 is the portion of 
the curve for which g y ^ 7r. It is represented by the principal 
value of the function, which is denoted by 

y — Cos" 1 x or y = Arc cos x. 

The Inverse Tangent. The inverse tangent function is 



y = tan" 



or y = arc tan x. 




H 1 H- 

-3 -2 -1 



7T/2 



-7T/2 



y»tan _1 x 



H 1 h- 

O l 2 3 



Fig. 8-12. 

It is represented by the graph in Fig. 8-12. We note that there are 
infinitely many valu.es of y for every value of x. 
The principal branch of the graph ofy = tan- 1 x is the portion 



7T 



7T 



for which — ~ <y <-~ * This is represented by the equation 
y = Tan -1 x or y = Arc tan x. 



160 



Trigonometric Functions; Inverse Functions 



Sec. 8-6 



The Inverse Cotangent, Secant, and Cosecant. The other inverse 
trigonometric functions are : 

y = cot -1 x or y = arc cot x, 
y = sec -1 x or y = arc sec #, 
2/ = esc -1 re or y = arc esc #. 




**x 



Fig. 8-13. 

We shall show only the graph of y = cot" 1 a?. See Fig. 8-13. The 
principal branch of this curve is given by < y < ir. 

Principal Values of the Inverse Cosecant and Secant. The selection 
of principal values of y = esc 1 x and y = sec 1 & is by no means uni- 
form among all authors. Some writers adopt the range between 
and 7r/2 for both functions when x is positive, and between —tt and 
— 7r/2 when x is negative. The authors prefer, however, to use the 
definitions 



Csc" 1 x 



and 



Sec -1 x = Cos -1 






We have, therefore, the following ranges of principal values of 
the inverse trigonometric functions : 

-\ SSiirisSf. 



- -5- Si Csc-1 * si 



^ Cos" 1 a: ^ 7r, 
< Cot" 1 x < ir, 
(Csc* 1 x * 0), ^ Sec" 1 x g tt (Sec" 1 z ^ y) 



y < Tan- 1 x < j > 



Sec, 8-6 Trigonometric Functions; Inverse Functions 161 

Example 8-3. Find the value of cos (Sin -1 -= J • 

Solution: Let Sin" 1 -= = 0. Then sin 6 = ■= > and cos = ± ^ • Since only the 
3 
principal value of sin -1 r is used, 6 is a first-quadrant angle and cos 6 cannot equal 

— -= • Hence, cos I Sin" 1 - j = - • 



Example 8-4. Find the value of sin [Tan" 1 ( - as)], 
x being a positive number. 

Solution: Let Tan" 1 ( - x) = 0. Then tan 
= — a;, and lies between - x/2 and 0. If angle 
is constructed in standard position, as shown 

in Fig. 8-14, then sin is found to be 

— x 



— x 



Vl + x 2 



Hence, sin [Tan -1 ( - x)] = 



Vl +z 2 




**X 



Example 8-5. Find Arc cos (cot 60°). 

Solution: Let Arc cos (cot 60°) == 0. Then cos = cot 60° = .5774, by Table II. 
Therefore, the value of is Arc cos .5774. Since is restricted to the interval from 
0° to 180°, must, in this case, be a first-quadrant angle. Hence, = Arc cos .5774 
= 54°44', and Arc cos (cot 60°) = 54°44'. 

EXERCISE 8-2 

In each of the problems from 1 to 6, find the inverse of the given function. 



l. y = 



2x_-5 
3 



A 4 _i7 

4 ^ = 15 X+ 6 



2.2/ = 
5. y = 



2x - 1 

6 
2x - 1 



Sx + 6 



3. y = mx + b. 
6. y = 2x - | • 



3 4 

In each of the problems from 7 to 22, find the value of the given expression. 



7. Sin" 1 



1 



11. Cot" 1 0, 
15. Tan 



3 



8. Tan-' 1. 
12. Sin- (- 1). 
16. Cos-- (^) 



9. Arc cos 



V3 



13. Cot 



-.VI. 



3 



10. Cos- 1 (- 1). 
14. Csc-» 1. 



19. Cos- 1 



(-&■ 



21. Sin" 1 ^-0.414). 



17. Tan" 1 ( - y/3). 18. Arc tan (- 1). 

20. Tan- 1 [sin (- ir/2)]. 

22. Tan- 1 (-1.414). 
In each of the problems from 23 to 37, evaluate the given expression. 

23. tan (sin" 1 j|) • 24. sin (sin- 1 1) • 25. sin (Tan-* ~) 

26. cot (| Sin- 1 ^) 27. cos ((tar 1 1) • 281 tan (Sin-* .6450). 



1 62 Trigonometric Functions; Inverse functions Sec. 8-6 

29. sin (Cos- 1 .9200). 30. tan (Cot-* 2). 31. sec (sin- 1 i) • 

32. sin (Cot-* i) • 33. cot (Cos" 1 1) • 34. sin (Cot~* |) • 

35. cos [Cot-* (-|)] # 36. cos (sin- 1 |) • 37. Sin- 1 (tan ^) • 

In each of the problems from 38 to 56 simplify the given expression, taking u as 
a positive number. In 38 to 40, 54, 56, u < t rr/2. In 44, 46, 48, 55, u <1. 
38. Cos- 1 (sin u). 39. Sin" 1 (- sin u). 

40. Sin- 1 (cos u). 41. esc f Sin- 1 - j • 

42. sin (Sin- 1 u). 43. tan (Tan" 1 v). 

44. sec (Sin- 1 y/T^t?). 45. cot (Tan- 1 * — \ • 

46. tan (Cot" 1 - = "j • 47. tan (Sin" 1 = \ • 

V \/l - ^ 2 ' ^ V^ 2 + 1/ 

48. esc (Sin- 1 y/l - u*). 49. tan (Cos" 1 J: ") • 

50. cos (Tan- 1 w). 51. sec (Cos -1 w). 

52. cot (Sin- 1 u). 53. sec (Sin- 1 u). 

54. Sin- 1 (sin u). 55. Cos" 1 (cos y/l - w 2 ). 

56. Cot- 1 f tan y U "j • 

V Vi + w 2 / 

In each of the problems from 57 to 65, draw the graph of the given function by 
changing from the inverse function to the direct function and using the period, 
amplitude, and phase angle of the function to assist in plotting. 

57. y = jr-sin-* x. 58. y = 2 tan -1 x. 

3 1 

59. y = - cos" 1 a;. 60. ?/ = x sin" 1 » — 1. 

61. y = g cos- 1 x + 2. 62. y = 4 tan" 1 x 4- 3. 

63. 2/ = — (Cos- 1 2x + 1). 64. y = 3 sin" 1 (2x + 1) - 2. 

7T 



65. 2/ = 4 tan- 1 (^-y^) 



66. Prove that sin (Cos- 1 u) £ 0, if < w £ 1. 

67. Prove that cos (Sin- 1 u) > £ 0, if < a ^ 1. 

68. Prove that cos (Tan- 1 u) > 0, if u £ 0. 

69. Prove that tan (Cos- 1 it) £ 0, if < u £ 1. 

70. Prove that Sin- 1 u + Sin- 1 (- u) = 0, if - 1 £ t* £ 1. 

71. Prove that Tan- 1 u -f Tan- 1 ( - w) = for all values of w. 

72. Prove that Tan- 1 u + Tan" 1 (1/u) = y i if u > 0. 

73. Prove that Cos" 1 w 4- Cos" 1 (-1*)=*-, if-l£u£l. 

74. Prove that Tan- 1 u - Tan~* v = Tan- 1 , tt ,"" - > if u > and v > 0. 

1 + uv 



V Linear Equations and Graphs 



9-1. SOLUTIONS OF SIMULTANEOUS EQUATIONS 

Often problems arise that involve two or more unknowns and as 
many equations. The solution of such a problem requires the deter- 
mination of numbers which simultaneously satisfy the given equa- 
tions. Equations for which we seek common solutions are referred 
to as a system of simultaneous equations. If the system has at least 
one solution, it is said to be consistent; otherwise, it is called 
inconsistent. 

Let us investigate the following pair of simultaneous linear 
equations : 

f a\x + buj = ci, 

(9-1) 

{ a 2 x + b 2 y = c 2 . 

It is desired to find all pairs of values of x and y which satisfy both 
equations, excluding from consideration the cases ai = &i = and 
On = b 2 = 0. 

We proceed by multiplying both sides of the first equation by b 2 
and both sides of the second by b u obtaining 

afax + biboy = cib 2 , 
a 2 bix + b 2 biy = c 2 b\. 

Subtracting the second equation of this pair from the first, we 
obtain 

(9-2) («i?>2 — a 2 b\)x = cib 2 — C2&i- 

If (ai&2 — a 2 b\) t* 0, [we find that 

c\b 2 — c 2 bi ^ 
aib 2 — a 2 b\ 

Thus, we see that we have exactly one value for x in any solution 
which may exist. 

163 



1 64 Linear Equations and Graphs Sec. 9—1 

By multiplying the original equations by a? and a u respectively, 
^nd subtracting the first equation thus obtained from the second 
one, we obtain 

(9-3) (aib 2 — d2bi)y = a x C2 — «2^i. 

If (a x b 2 — ^261) ** 0, we find that 

a\C2 — CL2C1 

y — • 

a x b 2 — «2&i 

Again, we have exactly one value for y in any solution which may 
exist. 

Hence, if (a x b 2 — &2&1) is not zero, we have at most one solution 
of the pair of given equations, namely, 

(9-4) *' x = Clh2 - C2bl 1 y = ai ° 2 ~~ a2Cl ■ 

d X b 2 — ^2^1 «1?>2 — #2^1 

Substitution of these values for a; and y in (9-1) shows that we 
really have a solution. The reader should verify this solution by mak- 
ing the substitutions. 

Consistent and Independent Equations. If {a x b 2 — (hbi) ^ 0, we 
have exactly one solution of (9-1), which is expressed by (9-4), 
and the given equations are called consistent and independent We 
may consider, as an example, the pair of equations 

( 2x + 3y = 24, 

1 bx - 2y = 22. 

Here a x b 2 - (hb x = (2) (-2) - (5) (3) = -19, and the values x = 6 
and y = 4 give a solution. In other words, when these values are 
substituted in the two given equations, both equations are satisfied. 
Thus, 

J 2(6) + 3(4) = 24, 

1 5(6) - 2(4) = 22. 

Cases with a x b 2 — chb x = 0. Let us now consider cases in which 
a x b 2 — a%b x = 0. If a x ¥" 0, division by a x gives 

b 2 = — 01. 
a x 

We then define k = — > and we have 

(9-5) a 2 = kau fa = ft&i. 

Now let #i = 0. Then, since 61 cannot be zero by our initial assump- 
tion, and since a 2 b x = 0, we have a 2 = 0. In this case, if we define 

k = ~ 1 we see that (9-5) again follows. Thus, a x b 2 - a 2 b x = has 
01 



Sec. 9-1 Linear Equations and Graphs 1 65 

been shown to mean that the left members of (9-1) are 
proportional. 

We note also that if a x b 2 — 0261 = 0, equations (9-2) and (9-3) 
become 

f • x = C1&2 — C2&1, 
(9-6) 

{ • y = a x c 2 - a 2 ci. 

It is clear that these equations cannot be satisfied by any pair of 
values of x and y, unless both right members are also zero. Accord- 
ingly, if we wish to determine whether or not the given system has 
a solution when a x b 2 - a 2 b 1 - 0, it becomes necessary to take into 
account the two cases of the right sides of (9-6) being zero or not 
zero. 

Consistent and Dependent Equations. In the first case referred 

to in the preceding paragraph, a x b 2 — chbi = 0, and c 1 b 2 — c 2 b t and 
a x c 2 — a 2 Ci are also equal to zero. Hence, substituting the values 
of b 2 and a^ from (9-5), we have 

{c\k — 02)61 = (cifc — c 2 )a\ = 0. 

Since a u bi are not both zero, it follows that c 2 = kc x . This means 
that one of the original equations is a constant multiple of the 
other, and any pair of values of x and y that satisfy one equation 
will also satisfy the other. Under this condition, the equations are 
said to be consistent and dependent. 

We have as an illustrative example the equations 

J 2x + 3y = 24, 

[ 4z + 6?/ = 48. 

Here a x b 2 - a 2 b x = (2) (6) — (4) (3) = 0. Also, the coefficients of 
the unknowns and the constant term in the second equation are 
multiples of the coefficients of the unknowns and the constant term 
in the first, and the multiplier is 2 for all three terms. Infinitely 
many solutions exist. Some of them are : x = 0, y = 8 ; x = 12, y = ; 

24 — 2t 
and x = t,y = — - — > where t is any number. 

d % 

Inconsistent Equations. Finally, let us suppose that in the orig- 
inal equations a x b^ — c^bi = and at least one of the numbers 
C\b 2 — c 2 bu a>ic 2 - CL2C1 is different from zero. Hence, no solution of 
(9-1) exists and the equations are inconsistent. This case is char- 
acterized by the condition that one of the numbers {c x k — (%) &i> 
(c x k — c 2 )di is not zero, in view of (9-5). It follows that c 2 ¥* kc x . 



1 66 Linear Equations and Graphs Sec. 9-1 

Hence, the multiplier for the left members of (9-1) does not apply 
po the constant terms. 

Consider, for example, the equations 

(2x + Zy = 24, 

[ 4x + &y = 7. 

Here the coefficients of the unknowns in the second equation are 
multiples of those in the first equation, and the multiplier is 2 for 
both terms; however, the multiplier for the constants is not the 
same as that for the other terms. Hence, these equations are 
inconsistent. 

9-2. ALGEBRAIC SOLUTION OF LINEAR EQUATIONS IN TWO UNKNOWNS 

To solve a consistent and independent system of two linear equa- 
tions in two unknowns, we reduce the system to one equation in one 
unknown by eliminating one of the unknowns. The following 
example will illustrate two commonly used methods for eliminating 
the unknowns. 

Example 9-1. Solve the equations 

2x+3y= 24, 

5x - 2y = 22. 

Solution: Since aM - a 2 bi = (2) (- 2) - (5) (3) = - 19, the equations are 
consistent and independent, and there is but one solution. If we use the method of 
elimination by addition or subtraction, the procedure is the same as that indicated 
in obtaining the solution (9-4) from equations (9-1). 

To eliminate y, multiply the first equation by 2 and the second by 3, in order to 
make the coefficients of y numerically equal in both equations. We thus obtain 

4a; + 6y = 48, 

15a; - 6y = 66. 
Adding, we get 19a; = 114. 

Solving for x, we have x = 6. 

Now, substitute 6 for x in the first of the original equations. Then 

32/ = 24 - 2a; = 24 - 12 = 12, 
or 

y =4. 
Alternate Solution: If we use the method of elimination by substitution, we begin 
by solving the first equation for y in terms of x. We thus get 

24 -2a; 



Sec. 9-3 Linear Equations and Graphs 1 67 

24 — 2x 

We then substitute — = for y in the second equation and obtain 



3 



Solving for x, we have 



or 



5, - 2 (?!=*;) = 22. 

15x - 2 (24 - 2x) = 66, 
15x - 48 + Ax = 66. 
19z = 114, 



Hence, 

and 

x = 6. 
Substituting 6 for #, as before, we find that y = 4. 

Example 9-2. A grocer has some coffee selling at 80 cents per pound and some 
at 90 cents per pound. How much of each must he use to get a mixture of 100 
pounds worth 86 cents per pound? 

Solution: Let x = number of pounds of 80-cent coffee, and y = number of pounds 
of 90-cent coffee. 
Then 

* + y = 100, 
and 

0.80a; + 0.902/ = 0.86 (100). 
Simplifying, we have 

x + y = 100, 

Sx + 9y = 860. 
These equations have the single solution x = 40, y = 60. 

9-3. LINEAR EQUATIONS IN THREE UNKNOWNS 

In the solution of a system of three equations in three unknowns, 
one method is to employ the following steps, which we do not justify 
here : 

1. Eliminate one of the unknowns from a pair of the equations; 
then eliminate this same unknown from another pair of the 
original equations. 

2. Solve the resulting two equations for the two remaining 
unknowns. 

3. Substitute the values found in step 2 in any one of the orig- 
inal equations to find the third unknown. 

Example 9-3. Solve the system of equations 

2x + 2y - 2=5, 

x — by + 2z = 1, 

Zx + y — 4« = — 1. 



168 



Linear Equations and Graphs 



Sec. 9-3 



Solution: Eliminate z from the first and second equations by addition to obtain 

5s 4- y = 11. 
Now eliminate z from the second and third given equations by addition to obtain 

5s — 9y = 1. 
We then consider the equations 

5s + y = 11, 

hx — 9y = 1. 

Solving these equations for s and y by subtraction, we have x = 2 and 2/ = 1. 
Substitution of these values in the first given equation gives 2=2. Hence, the 
solution of the given system is x = 2, y = 1, z = 2. 



EXERCISE 9-1 

In each of the problems from 1 to 30, solve the given system of equations. 
Check all solutions. 



L 



4. 



7. 



10. 



13. 



16. 



18. 



20. 



22. 



24. 



3s - 2y = 6, 
a; - 3y = 4. 

3s + 2y = 1, 
x - 2y = 5. 

3s + 2y = 4, 

2s - 3y = 3. 

s + 2?/ + 1 = 0, 
x - 4y + 2 = 0. 



11. 



3s - y = 7, 
2s 4- 2/ = 8. 

2s 4- 3y 4- 1 = 0, 
3s - 2/ + 7 = 0. 
x + 2y = 3, 
2s 4- 3$/ = 1. 
2s + Sy - 1 = 0, 
3s 4- 2/ + 3 = 0. 





f 1 


= 3, 


2,37 


3 , 1 


_1 


.-8 x+ 4 y= 9 - 


[2 X+ S V 


"""2 


3s - 2/ + 2« = 4, 






* + 2y - 3* = 1, 






2s - Sy + 2=2. 






s + 2y + 32 + 1 = 0, 






x - 42/ + 52 4- 1 = 0, 






2s + 62/ + 72 + 2 = 0. 






s 4- 22/ - 32 + 1 = 0, 






- x 4- 32/ - 42 - 5 = 0, 






2s 4- 62/ - 42 4- 3 = 0. 






3s + 2y + z =2, 






5s + 2/ 4-3=0, 






2x - 3?/ - 42 4- 5 = 0. 






3s - y 4- 42 = 2, 






4s 4- 4y 4- 42 = 5, 






>2s - y 4- 62 = 9. 







17. 



19. 



21. 



23. 



25. 



3. \y - 3s = 6, 

}s 4- 22/ = 3. 
6. (2s 4- 2y - 3 = 0, 

\ 5s 4- 32/ 4- 4 = 0. 
9. |3s + 2/ + 7 = 0, 

[4s 4- 82/ 4- 9 = 0. 
12. f 2s 4- 62/ - 7 = 0, 

J3s - 82/ 4- 9 = 0. 

15. f 2x 4- 42/ - 52 = 3, 
' 3s 4- y - 72 = 2, 
,4s 4- Sy - IO2 = 1. 

s 4- 22/ - 2=3, 

- s 4- 42/ 4- 22 = 1, 
3s 4- 2/ - 32 = 2. 

- 3s 4- 82/ 4- 92 - 3 = 0, 
2s 4- 32/ 4- 2-4=0, 
3s - 22/ - 22 - 4 = 0. 

3s - y 4- 52 = 0, 

s - 42 = 2, 

4s - 22/ - 32 4- 1 = 0. 

3s - 42/ 4- 22 = 3, 

2s 4- y =1, 

5s - 3i/ 4- 42 4- 5 = 0. 

(2s 4- Sy - 52 4- 2 = 0, 

52/ 4- 32 - 2 = 0, 

1 3 n 
*- 2/-2*~2 =0 ' 



Sec. 9-4 



Linear Equations and Graphs 



169 



26. 



3y + 2z = 4, 

2x - 2y + 3z = 3, 

I3x + 4a = 2. 



27. 



28. 



,2 2 1 



x y z 

x y z ' 



= 1. 



.-2+1=1 

x 2/ 3 



29. 



30. 



±+2+1 

I3(x +2/) -4(s 
1 4(3 + 2/) - 3(3 

3(3 + y) - 4(3 
4(3 - 2/) — 3(3 






For each of the following systems of equations, determine whether it is 
and independent, consistent and dependent, or inconsistent. 



31. 
34. 

37. 

40. 



2x - 3y = - 5, 32. 

43 - 6y = -- 3. 

2z + 4y = 3, 35. 

x + 22/ = 6. 
2z - 3?/ = 1, 38. 

4z + 6?/ = 2. 
- 3 + 3z/ = 2, 41. 

23 = 62/ 4- 14. 



(33 - 5y + 8=0, 
\ x + 82/ - 10 = 0. 
( 23 - 72/ + 1 = 0, 
\2ly -63 -3 =0. 
[ - 93 + 122/ = 3, 
33 + 42/ = 1. 



- 2/) = 5, 

- y) = 5. 

- y) = 5, 
+ 2/) = 7. 

consistent 
= 8, 



+ r=C 



lay + 63 = a&. 




63 + ay = a6c. 



9-4. GRAPHS OF LINEAR FUNCTIONS 

The discussion of rectangular coordinates in Section 2-1 set the 
stage for the pictorial representation of a function. By this repre- 
sentation of a function f(x), we mean the graph of the equation 
y = f(x). It consists of all points, and only those points, whose 
coordinates x and y satisfy the equation. 

In the same section we considered the graphing of lines parallel 
to the coordinate axes. The equation x = 3 was shown to represent 
a vertical line, that is, a line parallel to the y-axis which intersects 
the #-axis at the point (3, 0). This line thus includes all points 3 
units to the right of the y-axis. This example illustrates the fact 
that a linear equation in x alone represents a line that is parallel 
to the 2/-axis. Similarly, y = 2 was shown to be the equation of thfc 
horizontal line whiph is parallel to and 2 units above the #-axis. 
And this example illustrates the fact that an equation in y alone 
represents a line parallel to the z-axis. Furthermore, it is proved 
in analytic geometry that the graph of every first-degree, or linear, 
equation in x and y is a straight line; and, conversely, that every 
straight line is the crraDh of a linear equation. 



170 



Linear Equations and Graphs 



Sec. 9-4 



We shall proceed by first preparing a table of corresponding 
values in a given problem and then plotting the corresponding 
'points on the coordinate system to obtain the graph of the equation. 
The following illustrative examples will point the way toward an 
understanding of the procedure in the graphing of linear equations. 

Example 9-4. Graph the function 2x + 3. 

Solution: Let y = 2x + 3. Then assign any values for x ) substitute them in the 
equation, and obtain the corresponding values for y. The table and the graph are 
shown in Fig. 9-1. 

Since a straight line is definitely determined when two points are known, only two 
pairs of values of x and y are needed in graphing a linear equation. We can, how- 
ever, use three points in order to check our work. 



X 


y 





3 


1 


5 


-2 


-1 




*~x 



Fig. 9-1. 



Fig. 9-2. 



Example 9-5. Graph the equation 2x + 3y = 6. 
Solution: The equation may be solved for y in terms of x. Then 

y = -|a? +2. 
A table of corresponding values of x and y and the graph are shown in Fig. 9-2. 



9-5. INTERCEPTS 

In general, the points where a curve crosses the coordinate axes 
are the easiest to obtain. 

The ^-intercepts are the values of x at the points where the graph 
crosses the #-axis. Since y = on this axis, the ^-intercepts are the 
values of x that correspond to y = 0. Similarly, the ^/-intercepts are 
the values of y at which the graph crosses the y-axis. They are the 
values of y that correspond to x = 0. Hence, we have the following 
rule : 

To find the ^-intercepts, set y = in the equation and solve for x. 
To find the ^/-intercepts, set x = in the equation and solve for y. 



Sec. 9-6 



Linear Equations and Graphs 



171 



Example 9-6. Find the intercepts of the line 

2x + 3y = 6. 

Solution: To find the x-intercept, let y = 0. Then 2x = 6, and x = 3. 

To find the ^/-intercept, let x = 0. Then 3y = 6, and 2/ = 2. 

Note that the intercepts a; = 3 and y — 2 found in this solution correspond to 
the points (3, 0) and (0, 2), respectively, where the line in Fig. 9-2 crosses the 
coordinate axes. 



9-6. GRAPHICAL SOLUTION OF LINEAR EQUATIONS IN TWO UNKNOWNS 

In the graphical solution of two linear equations in two unknowns, 
the graphs of the two equations are drawn with reference to the same 
coordinate axes. Since the solution of two equations in x and y 
is a pair of values of x and y which satisfy both equations, the 
solution must represent graphically a point common to both lines 
represented by the equations. Hence, the values of x and y which 
satisfy both equations give the coordinates of the point of inter- 
section of the lines. We find, therefore, that the two lines intersect 
in a single point, are parallel, or are identical, according as the 
equations are consistent and independent, inconsistent, or consistent 
and dependent. 

Example 9-7. Solve graphically 

2x + 3y = 24, 

5x - 2y = 22. 

Solution: Tables of corresponding values for the two equations and also their 
graphs are shown in Fig. 9-3. 

It is seen from the graphs that the lines intersect at the point (6, 4). That x = 6, 
y = 4 gives the solution of the given equations may be checked by substitution. 

Y 



2x + dy = 24 bx - 2y = 22 




12 



x 


y 





-li 


22 




5 







**x 



Fig. 9-3. 



172 



Linear Equations and Graphs 
EXERCISE 9-2 



Sec. 9-6 



In each of problems from 1 to 9, graph the given function. In each case give the 
a:- and ^-intercepts. 



1. 4a: - dy = 1. 


2. ?/ = 2s - 8. 




3. y — x - 4. 


4. y = x + 5. 


5. 2/ = 3a\ 




6. 2/ = z. 


7. y = 3a: -f 4. 


8. y = a; - 3. 




9. y = 3a; — 5. 


Solve each of the following systems of equations graphically. 




10. f x - 3y = 1, 


11. 132/ -2a; =0, 


12. 


r 2?/ = x -3, 


[3a: - 2y = 0. 


\ x + y = 2. 




[2a: = ?/ + 3. 


13. (2?/ = a; + 3, 
12a: = 16 - 3?/. 


14. (3a; - y = 4, 
I 2/ - 3x = 1. 


15. 


[ - =8, 
x -y 

3 -4 




l2(x - y) 


16. (2a: - 0.-4, 
13a; + 2?/ = 12. 


17. !2x +4 = 5# - 3, 
13?/ -4 =4a; +2. 


18. 


.2a: - 3y = 12. 



10 



Determinants 



10-1. DETERMINANTS OF THE SECOND ORDER 

Let us consider the following system of two linear equations in 
two unknowns: , . 

f aix + b\y = ci, 
(10-1) 

The solution of these equations by the method of Section 9-1 is 
given by 

(10 2) x = — r r- > 2/ = — i — " i— * 

It is understood that t^fto — ffa&i "^ 0. 

We shall at this point introduce a more convenient way of writ- 
ing the expression iiib 2 — o-j&i. The notation which we select for 
this purpose will enable us to express also the numerators of (10-2) 
by means of the same symbol with the proper changes in letters. 

In choosing a symbol we shall select a form which will exhibit 
the numbers a u b u a*, b> 2 in the same relative positions as in (10-1) . 
Thus, we write 

Go 62. 
This arrangement of the four numbers in a square array, consist- 
ing of two rows and two columns, is then enclosed within vertical 
bars, as follows : 

oi 61 

a 2 bo 
This symbol represents a determinant of second order. 
Thus, we start with a square array, or matrix, such as 



\a 2 b 2 / 



173 



174 



Determinants 



Sec. 10-1 



We then associate with this array a number ai& 2 — & 2 &i, called its 
determinant, which is denoted in the following manner : 

The numbers a u b lf a 2 , b 2 are called elements of the determinant. 
The numbers ai and b 2 lie on the principal diagonal; the numbers 
a 2 and 61 lie on the secondary diagonal 

Note. We observe that the "expanded" value of the foregoing 
determinant is equal to the product of the elements on the principal 
diagonal minus the product of the elements on the secondary 
diagonal. It is interesting to note also that this value is the alge- 
braic sum of all possible products obtainable by taking one and 
only one element from each row and one element from each column. 
Each product is preceded by a plus sign or a minus sign, according 
to a rule to be stated in Section 10-2. 

Using the notation of determinants, we can write the solution 
(10-2) of (10-1) in the form 



(10-3) 



C\ 


h 


C2 


62 


O-l 


61 


0.2 


b 2 



ai 


Cl 


G2 


C2 


ai 


6l 


d2 


62 



We note that the value of each of the unknowns in (10-3) may be 
written as a fraction whose denominator is the determinant of the 
coefficients as they stand in (10-1), and whose numerator is the 
determinant formed from that of the denominator by replacing 
the column of coefficients of the unknown in question by the column 
of constant terms. 

Note. If a 2 = ka x and b 2 = kb lt where k is any number, then 

61 



a\ 



a 2 62 



= 0. 



In this case, the equations of the system (10-1) are inconsistent 
unless both numerators of the fractions in (10-2) are also equal to 
zero, that is, unless 



c\ 61 

C2 &2 



= and 



ai Ci 



«2 C\ 



= 0. 



Therefore, the equations of the system (10-1) represent distinct, 
parallel straight lines if c 2 ¥= kc u or they represent coincident lines 
if Co = kcu 



Sec. 10-2 



Determinants 



175 



Example 10-1. Solve the system of equations 

4x + Sy = 1, 

Sx - 2?/ = 22. 
Solution: Using determinants, we have 



-2-66 

-8-9 



Also, 



2/ = 



1 


3 


22 


-2 


4 


3 


3 


-2 


4 


1 


3 


22 


4 


3 


3 


-2 



68 



= 4. 



17 



88 



85 



= -5. 



- 8 - 9 - 17 



10-2. DETERMINANTS OF THE THIRD ORDER 

A determinant of the third order is a number designated by a 
square array of nine elements arranged in three rows and three 
columns and enclosed within vertical bars. An example is 

a\ bi c\ 



(10-4) 



D = 



0,2 62 C2 



«3 b-s ra 

The value, or expansion, of the determinant (10-4) is defined 
as the quantity 



(10-5) 



D =ai 



&2 C2 


-6l 


a 2 c 2 


+ Ci 


a 2 62 


63 C3 




«3 C3 




#3 63 



or as the quantity 

(10 0) D = tti62C3 — 01&3C2 + ^1^203 — &1C52C3 + Clfitefo — Ci&2«3, 

Here the products such as aib 2 Cs 9 aib 3 c 2 , and biases are known as 
the terms of the determinant. 

Minors and Cofactors. In any determinant the minor of a given 
element is the determinant of the array which remains after delet- 
ing all the elements that lie in the same row and in the same 
column as the given element. Thus, in (10-4) the minors of a u 61, 
Ci are, respectively, 



62 C2 




0,2 C2 




CL2 62 


63 £3 




as C3 




as 63 



176 



Determinants 



Sec. 10-2 



The cof actor of an element which lies in the ith row and ftth 
column is equal to the minor of that element if i -f k is even, and is 
equal to the negative of the minor if i + k is odd. That is, 

cof actor = (— l) i+k • minor. 

Thus, in the determinant in (10-4), the cof actor of a x equals the 
minor of ai, since a x lies in the first row and in the first column 
and i-ffc = l + l = 2, which is even. Similarly, the cof actor of 6i 
is the negative of the minor of b lf since i + fc = l + 2 = 3, which is 
odd. 

Often the following procedure may prove more convenient for 
finding the sign corresponding to a given element. Beginning with 
+ in the upper left-hand corner, change sign from place to place, 
moving horizontally or vertically, until the position for the element 
in question is reached. The schematic arrangement of signs cor- 
responding to the elements of a third-order determinant is thus 
as follows : 



+ 


— 


+ 


— 


+ 


— 


+ 


— 


+ 



Note that the sign for any position is independent of the path 
followed in arriving at that position. 

We shall designate the value of the cof actor of an element by the 
corresponding capital letter, and we shall use the subscript that 
occurs with the element itself. Thus, the cofactors of a u b u c x are, 
respectively, 



(10-7) 



Ai = 



b 2 
bs 



C2 



C'S 



Bi - - 



«2 C2 


, Ci = 


«2 &2 


a,3 C's 




«3 b' ti 



Hence, (10-5) for the expansion of the determinant may also be 
written as follows : 

(10-8) D = aUi + biBi + c x Ci. 

This sum is called the expansion of the determinant according to 
the elements of the first row. 

We observe at this point that the right member of (10-6) repre- 
sents all possible products, here 3 ! in number, that can be formed 
from the determinant in (10-4) by taking one and only one element 
from each row and each column. It follows also that the value 
of the determinant is the same, regardless of the row or column 



Sec. 10-3 



Determinants 



\77 



according to which the expansion is made. Thus, we may express 
the determinant as 

(10-9) D = biBi + b 2 B 2 + &3#3, 

or as 

(10-10) D = a 3 A 3 + 63B3 + C3C3. 

These equations represent the expansion according to the elements 
of the second column and according to the elements of the third 
row, respectively. 



Example 10-2. Expand the determinant 

2-5 3 



6 
- 1 



2 1 

7 4 



according to the elements of the first row and according to the elements of the 
third column. 

Solution: The expansion according to the elements of the first row is 



2 1 


+ 5 


6 1 


+ 3 


6 2 


7 4 




- 1 4 




- 1 7 



This reduces to 

2(8 - 7) + 5(24 + 1) + 3(42 + 2) = 259. 

Expanding according to the elements of the third column, we have 

= 3(42 + 2) - (14 - 5) + 4(4 + 30) = 259. 



(5 2 

-1 7 


-1- 


2 -5 
-1 7 


+ 4 


2 -5 
6 2 



10-3. PROPERTIES OF DETERMINANTS 

From the definition of the value of a determinant we may deduce 
the following important properties of determinants. These proper- 
ties supply us with more convenient methods for evaluating a 
determinant. 

Note. For a more complete discussion of these properties, the stu- 
dent is referred to any one of the various treatises on determinants 
or to texts on the theory of equations or on solid analytic geometry, 
where he will also find proofs which apply to determinants of 
any order. 

The properties listed here will be employed in examples that fol- 
low, and their usefulness in simplifying determinants will be 
illustrated. 



178 



Determinants 



Sec. 10-3 



Property 1. The value of a determinant is not changed if its rows 
and columns are interchanged. 

Property 2. If all the elements of a row, or of a column, are multi- 
plied by the same number, the value of the determinant is 
multiplied by that number. For example, 



ka\ b\ 


= fc 


a x 61 


ka,2 62 




0,2 62 



Property 3. If two rows, or two columns, of a determinant are 
identical or proportional, the value of the determinant is zero. 
For example, let the first two columns be identical, as in the 
determinant 

a\ ai c\ 

D = 0,2 CL2 C2 

03 03 c$ 

Then, expanding according to the elements of the third column, 
we have 



D = a 



02 


02 


- C 2 


01 


Ol 


+ C 3 


01 


Ol 


O3 


03 




03 


03 




02 


02 



= 0. 



Oi 


61 


Ci 




a 2 


b 2 


c 2 


- 


0.3 


63 


C3 





Property 4. The value of a determinant is not changed if we add 
to the elements of any column (row) any arbitrary multiple of the 
elements of any other given column (row) . 

For example, 

rti + nb\ b\ C\ 
«2 + Tib2 62 C2 • 

«3 + W&3 b 3 C 3 

The proof follows. Expanding according to the elements of the 
first column, we find that 
01 + nbi 61 c\ 

tt2 + W&2 &2 C2 

03 + nbz 63 C3 
The last determinant vanishes, since two columns are identical. 



Ol 


61 


C\ 




02 


b 2 


C2 


+ n 


03 


63 


C3 





61 


61 


C\ 


b 2 


62 


C2 


h 


h 


C3 



Sec. 10-3 



Determinants 



179 



Example 10-3. Evaluate the determinant 

4 3-1 



D = 



5 1 
2 4 



Solution: By adding 2 times the elements of the first row to the elements of the 
second row, we obtain 



4 


3 


- 1 




4 


3 


- 1 


5 


1 


2 


= 


13 


7 





2 


4 


3 




2 


4 


3 



If now we add 3 times the elements of the first row to the third row, the determinant 
becomes 

4 3-1 



13 7 

14 13 



Expanding according to the elements of the third column, we have 

13 7 



(-D 



14 13 



= -71. 



In this example, we first converted the given determinant to one in which all but 
one of the elements of the third column are zero. For the final expansion, the given 
determinant was thus reduced to a determinant of the second order, and its value 
was easily found. 



Example 10-4. Without expanding the determinant, show that x = — 2 satisfies 
the equation 



3x 2 x z - x 
3 1 7 

6-4 1 



= 0. 



Solution: Substituting — 2 for x in the determinant, we have 

12 -8 2 

3 1 7 

6 -4 1 

This equals zero, since the first and third rows are proportional. Hence, the 
equation is satisfied by x = — 2. 



180 



Determinants 
EXERCISE 10-1 



Sec. 10-3 



In each of the problems from 1 to 12, evaluate the given determinant. 



2 -5 

3 -5 

3-5 
6-10 

2 
- 1 3 



2 7 

3 -5 



3, 



3 4-5 



6. 



10. 



3 


8 


7. 


6 


-6 





X 


y 


1 




Xi 


Vi 


1 


• 11. 


x 2 


2/2 


1 





7 2 
14 4 

1 -3 2 

2 1 4 
-2 1-3 

2 4-5 

-6-1 4 

4 8-9 



12. 



9 7 

8 14 

2 2 
3-2-1 

2 1 4 

2 2 6 

1 -6 3 

5 7 15 



♦ In each of the problems from 13 to 19, solve the given system of simultaneous 
equations by means of determinants. Cheek all solutions by substitution in the 
equations. 



13. f x + 3y = 5, 

\2x - 4?/ = 7. 



14. \2x - y = 1, 

Ux +2y =4. 



15. 



Sx + 2y = 5, 
2x - Sy = 5. 



16. If D represents the determinant in Problem 10, show that D = is the equation 
of the straight line through the points (x lt ?/,) and (x 2 , 2/2). 

17. Find the ^-intercept and the ^-intercept of the line whose equation is 

x y 1 

3 4 1 =0. 

2-3 1 

18. Solve graphically the following system of equations: 



= 0. 



19. Find the coordinates of the vertices of the triangle whose sides are the straight 
lines 3-0+2=0, 2x + 9y + 15 = 0, and 7x + Ay - 30 = 0. 

20. Find the vertices of the parallelogram formed by the following lines: 



= 0. 



X 


y 1 




2 


1 


= 0, 


1/2 


1 1 





x y 


1 


1 


1 


1/2 


1 



x y 


1 




2 1 


1 


= 0, 


-3 2 


1 





X 


y 1 




x y 1 




6 


- 1 1 


= 0, 


1 3 1 


= 0, 


1 


1 




2 4 1 





X 


y 


1 


5 


1 


1 


1 


-3 


1 



Sec. 10-4 



Determinants 



181 



10-4. SOLUTION OF THREE SIMULTANEOUS LINEAR EQUATIONS IN THREE 
UNKNOWNS 

Let us consider the following system of linear equations : 

a x x + hy + az = d\ } 



(10-11) 



a 2 x + b 2 y + c 2 z = d 2 , 



, d3X + foy + c 3 z = dz. 
The determinant of the coefficients of the unknowns is 

a\ b\ c\ 



D = 



a 2 



c 2 



(iz bs C3 

For the solution of such a system, we employ the following theorem, 
which is known is Cramer's Rule. 

Theorem. If the determinant D of the coefficients of the system 
is not equal to zero, the system has just one solution. In this solu- 
tion, the value of any unknown is equal to a fraction whose denom- 
inator is D and whose numerator is obtained from D by replacing 
the column of coefficients of the unknown in question by the column 
of constants d u d 2 , and d> 6 . 

Proof. Let the numerators of the fractions for x, y, z be denoted 
by D u D 2 , D 3 , respectively. We proceed to show that if equations 
(10-11) are to be satisfied, then 

(10-12) Dx = D u Dy = D 2> Dz = D 3 . 

Specifically, the equation for x will have the form 

ai b\ c\ 



(10-13) 



a 2 



as 



b 2 



c 2 



C'S 



di 


bi ct 


d 2 


62 C2 


d 3 


63 Cz 



Similar equations may be written for y and z. 

To find x, the first equation of the system in (10-11) is multi- 
plied by the cof actor A u the second by A 2 , and the third by A 3 . 
After adding and collecting terms, we obtain 

(10-14) (aiAi + a 2 A 2 + a%A$)x + (biAi + b 2 A 2 + foAs)y 

+ (ciAi + c 2 A 2 + c$Az)z = d\A\ + d 2 A 2 + d^As. 

The coefficient of x in (10-14) is the expanded value of D accord- 
ing to the elements of the first column. The coefficient of y is 

biAi + b 2 A 2 + 63A3; 



182 



Determinants 



Sec. 10-4 



this is equal to 



61 


h 


C\ 


62 


b 2 


C2 


&3 


h 


C3 



which equals zero, since two columns are identical. Similarly, the 
coefficient of z is zero. Hence, we have shown that the left side of 
(10-14) is the expanded form of the left side of (10-13) and that 
the two are therefore the same. 

The right side of (10-14) is cM-i + d 2 A 2 + d 3 A 3 , which is the 
expansion of the determinant 

d\ 61 c\ 

d>2 &2 C2 

dz b 3 c-3 

This determinant may be obtained from D by replacing the coeffi- 
cients of x by the column of constants ; that is, it is the expansion 
of the determinant that we have called Di. Hence, the equation 
Dx = D x in (10-12) is established. By a similar procedure we can 
show that the equations Dy = D 2 and Dz = D 6 are also valid. 
If D ¥* 0, the value of x is given by the equation 



(10-15) 



di 


h 


Cl 


d 2 


62 


C2 


d 3 


bs 


C3 



X = 



tti 61 Ci 

a2 62 C2 



tt3 63 C3 

Similar values may be found for y and z. The proof is complete. 

The proof of (10-12) is valid whether D ^ or D = 0. If Z) ^ 0, 
the equations of the system (10-11) are consistent and have only 
one solution, which is of the form (10-15) ; that is, 



(10-16) 






y = 



D2 

D 



z = 



D3 
D ' 



If D = 0, and any one or more of the other determinants, D u D 2 , 
Z) 3 , is not zero, the given system of equations has no solution and is 
inconsistent. 



Sec- 10-5 



Determinants 



183 



If D = 0, and all the other determinants are zero, the equations 
of the system (10-11) may be consistent or inconsistent. If they 
are consistent, there are infinitely many solutions. This case will be 
treated in Section 10-5. 

The following example will illustrate the case for which D ¥=0. 

Example 10-5. Solve the system of equations 

x - y + z = 1, 

x + y - 2z = 3, 

2x - y -f dz = 4. 

Solution: Here 

1-11 1-11 

D = 1 1 -2=5; Di = 3 1 - 2 = 11; 
D 2 = 1 3 -2 =8: D s = 1 1 3=2. 



1 


-1 


1 






1 


- 1 


1 


1 


1 


-2 


= 5; 


Di = 


3 




. -2 


2 


- 1 


3 






4 


- 1 


3 


1 


1 


1 






1 


- 


I 1 


1 


3 


-2 


= 8; 


Z>3 = 


1 




L 3 


2 


4 


3 






2 


— 


L 4 



D 8 Z) 2 

Hence, x=^=~>y=z-j£~-iz=-^=-' If we check by substitution, we 






11 
5 



D 



D 



find that these values satisfy the given equations. 



10-5. SYSTEMS OF THREE LINEAR EQUATIONS IN THREE UNKNOWNS WHEN 
D = 

We note that when D = 0, the system (10-11) will not have a 
solution if any one of the other determinants D u D 2 , D 3 is different 
from zero. Suppose that a solution is given by 

x = r, y = 8, z = t. 
Then the equations (10-12) become 

r • = Z>i, 8 • = D 2 , * • = D 3 . 

It follows that Z?i = 0, D 2 = 0, Z) 3 = 0. 

The following example will illustrate the case of consistent equa- 
tions where D = and D x = D 2 = D s = 0. The equations are said to 
be dependent, and they have infinitely many solutions. The student 
should construct an example to show that the equations (10-11) 
may be inconsistent when D = 0, even though D x = D 2 = D 3 = 0. 



184 



Determinants 



Sec. 10-5 



Example 10-6. Solve the system of equations 

x + y - z =3, 

2x - 2/ + 3z = 1, 
4z - 2y + 6z = 2. 



Solution: Here 



Z) = 



1 1-1 
2-1 3 

4-2 6 



This equals zero because the second and third rows are proportional. But, we also 
find that 

1 



D x = 



3 1-1 
1-1 3 
2-2 6 



= 0, D 2 = 



13-1 
2 1 3 

4 2 6 



= 0, 7) 8 = 



1 3 
- 1 1 
-2 2 



= 0. 



In this case the given equations have a solution. In fact, the second and third 
equations are proportional, and so either of these can be solved together with the 
first equation for two of the unknowns in terms of the third. For example, 



x = • 



4 - 2z 



y =■ 



5 +5z 



3 ' y ~ 3 

Thus, we have a single value of x and a single value of y for every value of z. 
However, there are infinitely many values of z, and therefore infinitely many 
solutions of the given equations exist. 

10-6. HOMOGENEOUS EQUATIONS 

The system (10-11) is homogeneous if dx = 0, d 2 — 0, and d 8 = 0. 
Such a system always has the trivial solution x = y — z = 0. When 
d x = d 2 = rf 3 = 0, it is seen that J5 X = D 2 = Z) 3 = 0, for each of these 
determinants has zero for every element in one column. If D ¥= 0, 
it follows from (10-16) that we can have but one solution, which 
is given by 

Di _ n .. _ D 2 _ n . _ Z) 8 



^ = ^ = 0, 



» = -5= = ' 



= 



D 



0. 



Hence, if the given system is to have a solution besides the trivial 
solution, D must equal zero. It may be shown that, if D = 0, non- 
trivial solutions always exist. 

Example 10-7. Solve the system of equations 

x - y + z = 0, 

2z - 3y + 42! = 0, 

5x - 2y - z = 0. 



See. 10-7 



Determinants 



185 



Solution: The determinant D is 



1 


- 1 


1 




2 


-3 


4 


= 


5 


-2 


- 1 





0-1 

-1-3 1 

3 - 2 - 3 



= 0. 



Therefore, nontrivial solutions exist. To find these solutions, we proceed as follows: 
Transpose z in each of the first two equations, and solve for x and y in terms of z. 
Then we have 



D = 



Hence, 



1 

2 


- 1 
-3 


= -1, Dt = 
Di 


-z - 1 
-4z -3 

= 2, and 


= - 2, Z> 2 = 


1 

2 


— z 
-42 



= -22. 



Substitution shows that these values also satisfy the third equation. The given 
system therefore has infinitely many solutions, and the values of x, y, and z are re- 
lated by the equations x = z and y = 2z. 



EXERCISE 10-2 

In each of the problems from 1 to 6, solve the given system of simultaneous 
equations by means of determinants. Check all solutions by substitution in the 
equations. 

1. {Sx + y - z = 11, 2. ( x- y-2z=-l, 

hx -2y = 0, 

12s - 4?/ -f z=3. 
2x - ?y - 3z = 7, 6. 

x 4- 2y - z = 10, 
3s - 3*/ + 22 = - 7. 
For each of the following systems find at least one nontrivial solution, or show 
that there is no nontrivial solution. 



\3x + y - 2 = 11, 2. 

z + 3?/ - 2 = 13, 
[ a? + y - 3z = 11. 
\2x + y +32 = - 10, 5. 

2s - 2// + 2=2, 
[6x + 2// -22 =5. 



a? - 2/ + 62 = 7, 
\2x + 3y + 62 = 0, 
( £ + 2y + 92 = 3. 

[2j - 2/ = 3, 
2x -32 = - 1, 
(32 - y = 2. 



7. I 3 - 2i/ + 22 = 0, 
J 2s - by +2=0, 
[4z — 11?/ — 2=0. 



8. 



2s - 3?/ + 42 = 0, 

x +3y - z = 0, 

7s + 3y + 52 = 0. 



9. 



z - 2y + 32 = 0, 
2x - ?/ + 42 = 0, 
3.r + 2/ - 2=0. 



10-7. SUM AND PRODUCT OF DETERMINANTS 

.Closely related to Property 4 of Section 10-3 is a theorem con- 
cerning the sum of determinants. We shall illustrate the theorem 
for splitting the elements of a given column into two parts by 
means of the following equality for third-order determinants : 



ai + 61 ci ^ 

Q>2 ~f" &2 C2 d,2 
Q>3 + &3 C3 (#3 



ci\ C\ d\ 




Q>2 C2 G?2 


+ 


03 C3 dz 





61 Ci rfi 

62 C2 C?2 

63 C3 C?3 



186 



Determinants 



Sec. 10-7 



This can be shown to be true by expanding the three determi- 
nants according to the elements in the first column and noting that 
in the expansion the minors are the same for all three determinants. 



Example 10-8. Show that 



2 


4 5 




1 


4 5 




3 


4 5 


1 


- 1 


+ 


4 


- 1 


= 


5 


- 1 


3 


7 6 




- 1 


7 6 




2 


7 6 



Solution: Expanding each of the determinants on the left side of the equation 
according to the elements in the first column, we have 



- 1 

7 6 


- 


4 5 

7 6 


+ 3 


_ 


4 5 
1 


+ 




- 1 

7 G 


-4 


4 

7 


5 
6 


- 


4 5 
- 1 






= 3 


- 1 

7 



G 


-5 


4 

7 


5 
6 


+ 2 


4 
- 1 


5 



• 









ai 


bi 




C\ d\ 


= 






+ 






a 2 


b 2 




0,2 &2 



This result is the expansion of the determinant on the right in the given equation 
according to the elements of the first column. Computing the value of each determi- 
nant in the given equation, we see that each side reduces to 47. 

A similar theorem for splitting the elements of a given row is illustrated by the 
following example : 

ai + Ci 61 + di 

CL2 &2 

This can be shown to be valid by expanding according to the elements of the 
first row. 
Thus, consider the determinant 

4 5 

1 7 

This may be written as a sum in various ways. Examples arc 



2 3 


+ 


2 2 


and 


4 5 


+ 


4 5 


1 7 




1 7 




-2 2 




3 5 



We shall state without proof the rule for the product of two determinants of 
the same order. The product is equal to a determinant of like order in which the 
element of the ith row and kth. column is the sum of the products of the elements 
of the ith row of the first determinant and the corresponding elements of the kth 
column of the second determinant. For example, for second-order determinants, 
we may write 

1-5 +2-7 1-6 +2-8 



1 2 
3 4 



5 6 

7 8 



3-5 + 4*7 3«6+4'8 



Sec. 10-7 



Determinants 



187 



That the product of the values of the two determinants on the left equals the 
value of the determinant on the right is checked by expanding. Thus, the desired 
equality becomes 

(4 - 6) • (40 - 42) = (950 - 946), 
or 

(-2). (-2) =4. 

To illustrate the multiplication of two third-order determinants, we have that 

1 7 6 

- 3 9 3 

5-42 



3 


2 5 




1 


- 1 2 


• 


4 


6 





is equal to 
3 • 1 + 2( - 3) + 5 • 5 



3-6+ 2-3+5-2 



3-7+2-9+ 5( - 4) 
1- 1 +(- 1) (-3) +2-5 1-7 + (-1)9+ 2(- 4) 1 • 6 + (- 1)3 + 2 • 2 
4-l+6(-3)+0-5 4-7+6-9+0(-4) 4-6+6-3+0-2 

This reduces to 



22 


19 


34 


14 


- 10 


7 


14 


82 


42 



which is equal to — 630. Also, computing the values of the given factors, we have 
30(- 21), which equals - 630. 



EXERCISE 10-3 

In each of the first three problems, combine the given determinants into a single 
determinant, and evaluate the result. 



1. 



2 




2 




2 




2 


- 


7 










-3 - 1 


7 


+ 


-3 


- 1 


7 


• 






4 8 


- 10 




4 


8 


- 10 








1 2 


2 




1 4 


2 










2-3 


1 


- 


2 2 


1 


• 








4 8 - 


1 




4 19 - 


- 1 










2 4 


-5 




2 


4 


-5 




2 


4 


-5 


-6 - 1 


4 


+ 


- 6 


- 1 


4 


+ 


- 6 


- 1 


4 


1 2 




12 




4 


j 


5 


-6 






- 1 


3 


- 16 



188 



Determinants 



Sec. 10-7 



In each of the following problems, find the product of the determinants without 
evaluating the individual factors. 





3 1 




3 


1 


4. 




. 






1 -2 




1 5 


11 21 12 3 




6. 1 • ] 


. 




2 3| 1 


3 


1 





/I 12 3 -1|\ I 
\| — 1 4 5 l|/ I - 

(V - 2 . 4 %( 

\| 9 3 6 7|/ \ 



2 6 

3 8 



13 21 
5 4 



2 

- 1 
3 

1 



-3 2 
1 4 



I)' 



-2 

4 

4 

10 



11 



Complex Numbers 



11-1. THE COMPLEX NUMBER SYSTEM 

There are many problems that cannot be solved by the use of 
real numbers alone. We observe, for example, that the equation 
x 1 + 1 = has no real root, since x 1 can never be negative if # is a 
real number. In order to provide solutions to such equations, a new 
system of numbers, called the complex number system, was intro- 
duced. Later in this book, we shall find many instances of solutions 
involving complex numbers. 

We shall now define a complex number as an ordered pair of real 
numbers, which we denote by (a,b). If the numbers a and b are 
regarded as the Cartesian coordinates of a point in a rectangular 
coordinate system, we have a one-to-one correspondence between 
the set of complex numbers and the set of points in a plane. The 
plane is called the complex plane. Two complex numbers (a, b) 
and (c, d) are equal if and only if they correspond to the same 
point, that is, if and only if a = c and b = d. 

Addition, subtraction, and multiplication of complex numbers 
are defined as follows : 

(11-1) (a, b) + (c, d) = (a + c, b + d), 

(11-2) (a, b) - (c, d) = (a - c, 6 - d), 

(11-3) (a, b) • (c, d) = (ac - 6d, ad + 6c). 

For example, 

(1, 3) + (5, 2) - (G, 5), 
(1, 3) - (5, 2) = (- 4, 1), 
(1,3). (5,2) = (-1,17). 

We also define the following special complex numbers: 
(11-4) = (0,0), 

(11-5) 1 = (1, 0), 

(11-6) % = (0, 1). 

189 



1 90 Complex Numbers Sec. 11-1 

The complex number serves as a zero of the complex number 
system, while 1 serves as a unit, in accordance with the following 
properties : 

+ (a, b) = (a, b) + = (a, b), 

• (a, 6) = (a, 6) • = 0, 

1 • (a, 6) = (a, 6) • 1 = (a, 6). 

The so-called imaginary unit i= (0, 1) will be discussed in more 
detail in Section 11-2. 

If A; is a real number, we define 

(11-7) * • (a, b) = (fc, 0) • (a, 6) = (fca, kb). 

Also, we define 

(11-8) - (a, 6) = (- 1) • (a, b) = (- a, - 6). 

Since (a, 6) + (—a, —6) = (0, 0), the complex number —(a, b) is 
called the negative of (a, 6). 

The Reciprocal of (a,b). If (a, b) ¥= 0, then it has a reciprocal 
(#, #) such that 

(a, 6) • (x, ?/) = 1. 
Furthermore, the reciprocal is given by 

+ 6 2 ' a 2 + b 2 ) 
By (11-3) and (11-5), the equation (a,, b) *\x,y) =1 may be 
written in the form 

(ax — by, ay + bx) = (1, 0). 
Since a 2 + b 2 ¥= 0, we may determine x and 2/ by solving the simul- 
taneous equations 

[ ax — by = 1, 

( for + a?/ = 0. 
The values of x and 7/ can be foundry any of the methods taken up 
in Section 9-1. The solution is 

_ a __ b 

X ~ a 2 + b 2 ' y ~ ~ a 2 + b*l 
We have, therefore, verified (11-9) . 

Division of (a,b) by (c, d). If (c, d) ¥= 0, division can be defined 
as follows : 

(a, b) -v- (c, d) = (a, 6) • (u, «), 
where (w, t;) is the reciprocal of (c, d). By (11-9), we have 

(11-10) (a, b) + (c, d) = (a, 6) (^-^ , - ^5) 



(11-9) (X|V) =(«_,_ _^_) 



+ cP c 2 + 
— ad -\- b 
c 2 + d 2 ' c 2 + d 2 



_ /ac + bd — ad + &c\ 
~ \c 2 + d 2 ' c 2 + d 2 ) 



Sec. 1 1 -2 Complex Numbers 1 9 1 

Forexamp.e, (5, 13) * (3, - 2) = (- » , {§) . 

This result can also be verified by the method in Section 11-3. 

11-2. THE STANDARD NOTATION FOR COMPLEX NUMBERS 

The special number i = (0, 1), defined by (11-6), has the follow- 
ing property : 

(11-11) t* = (0, 1) • (0, 1) = (- 1,0) = - (1, 0) = - 1. 

We shall now show that (a, 6) and the binomial form a + bi are 
equivalent, or that 

(11-12) (a, b) = a + bi, 

in which a + bi means al 4- bi. By (11-5), (11-6), and (11-7), 

a\ + bi = a(l, 0) + 6(0, 1) = (a, 0) + (0, 6). 

Finally, by (11-1), we have 

(a, 0)4- (0,6) = (a,6). 

Hence, (11-12) is established. 

Real and Imaginary Parts of Complex Numbers. We call a the 
real part and & the imaginary part of the complex number a 4- bi. 
If a = and 6^0, a + bi reduces to bi, which is called a pwre 
imaginary number. If 6 = 0, the complex number a + bi, or al 4- bi, 
reduces to the complex number al, which may be identified with 
the real number a. The complex numbers, then, include both the 
real numbers and the pure imaginary numbers as special cases. 

Illustrations of various classes of numbers follow : 
Some real numbers are —2, 5, and y/S. 
Some pure imaginary numbers are Si, and — y/5i. 
Some complex numbers are 2 4- Si, and y/S — i. 

Note that the numbers —2, 5, \/S, Si, and — y/Ei may be put into 
the standard form a 4- bi and written, respectively, as the complex 
numbers -2 4- Oi, 5 4- Oi, \/S 4- Oi, 4- Si, and — \/5i. Since 
= 04- Oi, which may be written briefly as 0, we shall drop the use 
of bold-face 0; similarly for bold-face 1. 

Conjugate Complex Numbers. The conjugate of a complex num- 
ber a 4- bi is defined as a — bi. Likewise, a 4- bi is the conjugate of 
a — bi. Some pairs of conjugate complex numbers follow: 2i, —2i; 
3 4- 5i, S — 5i; and x 4- 2yi, x — 2yi. A real number is its own 
conjugate. 



192 Complex Numbers Sec. 11-2 

Powers of t. It is readily seen that i 3 = i 2 • i = — i, i* = i 2 • i 2 = 1, 
i& = i* . i = ^ £6 = z *4 . z -2 = _^ an( j so on Therefore, successive posi- 
tive integral powers of i have only four different values, namely, 
i, —1, — i, and 1 ; these four values are repeated in regular order. 
Hence, if n is any positive integer, we have in general 



i 4n 


= 


i 4 


•=. 


1, 




i4n+l 


= 


i, 








i4n+2 


= 


i 2 


= 


— 


i, 


f4»+3 


= 


i* 


= 


— 


i. 



These relationships afford a simple method for evaluating powers 
of i, as shown by the following illustrations : i 7 = i 4+3 — i 3 = — i ; 
i 38 = i 4 * 9+2 = i 2 = -1 ; and i 103 = i*- 26 *- 1 = t. 

11-3. OPERATIONS ON COMPLEX NUMBERS IN STANDARD FORM 

From the definitions given in Sections 11-1 and 11-2, it follows 
that a 4- bi and c + di can be added, subtracted, multiplied, and 
divided as if they were real binomials, except that, where i 1 appears, 
it is replaced by —1. 

Algebraic Addition and Subtraction. Addition of two complex 
numbers is effected by adding their real and imaginary parts sepa- 
rately ; and subtraction is performed by subtracting their real and 
imaginary parts separately. Thus, in accordance with (11-1) and 
(11-2), 

(11-la) (a + bi) + (c + di) = (a + c) + (b + d) i, 

and 
(ll-2a) (a + bi) - (c + di) = (a - c) + (b - d) i. 

For example, 

(3 + 20 + (4 - 5i) = (3 + 4) + (2 - 5)i = 7 - 3t, 
and 

(3 + 20 - (4 - 50 = (3 - 4) + (2 + 5)i = - 1 + 7i. 
We note that the sum of conjugate complex numbers is a real num- 
ber, because (a + bi) + (a — bi) = 2a. Also, the difference of two 
conjugate complex numbers is a pure imaginary number, because 
(a + bi) - (a - bi) = 2bi. 

Algebraic Multiplication. To find the product of two complex 
numbers, multiply them according to the rules of algebra, and 
replace i 2 by -1 in the result. Thus 

(a + 60 (c + di) = ac + adi + bci + bdi 2 . 
Replacing i 2 by —1, we have, in agreement with (11-3), 
(ll-3a) (a + bi) (c + di) = (ac - bd) + (ad + bc)i. 



Sec. 11-3 Complex Numbers 193 

In many respects the notation a + bi is more convenient than 
(a, b) . In particular, the former notation makes it easier to remem- 
ber how to multiply two complex numbers. For example, 

(3 + 2i) (4 - 5z) = (12 + 10) + (- 15 + 8)t = 22 - 7t. 

The student should note that the product of two conjugate complex 
numbers is a non-negative real number, because (a + bi) (a — bi) = 

a 2 + b 2 . 

Algebraic Division. To obtain the quotient of two complex num- 
bers, multiply the numerator and the denominator by the conjugate 
of the denominator. Thus, if c + di ¥= 0, 

a + bi __ a + bi c — di __ ac — adi + bci — bdi 2 

c + di ~" c + di c — di "" c 2 — d 2 i 2 

__ (ac + fed) + (be — ad)z 

~ c 2 + d 2 

Therefore 

' a + 6i _ ac + 6d , 6c — ad . 

^Tdi "^Td 2 + 1?+W %m 

The right member is, of the form A + Bi, and this equation agrees 
with (11-10). 

Example 11-1. Reduce - to the form a -f bi. 
Solution: The conjugate of i is — i. Then 

i i ( — t) - i 2 +1 

Example 11-2. Find the value of 5 + 13i divided by 3 - 2i. 

5 4- 13t 
Solution: Represent the division as -r — -yr > and multiply the numerator and 

the denominator by 3 + 2i. We get 

5 + 13i 3 +2i (15 - 26) + (39 + 10)t _ 11 49 . 

3 - 2i ' 3 + 2i "" 9+4 13 + 13 ^ 



EXERCISE 11-1 

In each of the problems from 1 to 12, express the given quantity in the form 
a + bi and give its conjugate. In working these problems, note that y/ — a 2 
= y/a? y/^~l = | o 1 1. 



1. V- 16. 


2. V~25. 


3. - V^. 


4. y/-X*. 

7. 3 a/2 + 3 y/~2. 


5. - V- 3(K7 2 . 

8. 4 - 3 V- 16. 


6. V- *V« 

9. 1 + 2 V" 17 ^. 



10. v/^ 2 - V- * 2 . 11. V15 4- y/- 64a 3 6. 12. 3 + V- 32a 2 6 3 . 



1 94 Complex Numbers Sec. 1 1-3 

In each of the problems from 13 to 26, compute the value of the given expression. 
13. i". 14. i 12 . 15. (- i) 13 . 16. - (- i) 17 . 

17. i«. 18. -x 70 19. (-i) 236 . 20. (-i) 602 . 

21. i 3 i 29 . 22. t 37 i 183 . 23. i 14 - (- i) 18 . 24. i 10 + i 20 + i 30 . 

25. i 25 - i 50 4- i 76 - i 100 . 26. i 10 4- i 100 4- z 1000 4- i 10000 . 

In each of the problems from 27 to 36, find the values of x and y which satisfy the 
given equation. 

27. (2jc, 3?/) = (3, 1). 28. (2x, Sy) = (8, 9). 

29. (3x, 5?/) = (18, - 25). 30. (x + ?/), 3x + 1) = (x, 1). 

31. (3* + 2y,x + 5?/) = (2?/ + 3, - 19). 
32. z + (y - x)i = 1 + 3i. 33. 3z - 6 - (5 - 2t/)i = 0. 

34. 2a; - 4yi = 6 - 2si. 35. 3x - 7 = (4 - Sy)i. 

36. 3xt - 2yi + 8x + 5y - 12 = 2x + 2?/i + 5?/ + 6 + (x - 2)i. 

In each of the problems from 37 to 78, perform the indicated operations and 
reduce to the form a -f- hi. 

37. (2, - 3) + (5, 6). 38. (1, - 1) + (3, 2). 39. (1, V§) + Q > - ^) • 

40. (1, - 7) - (7, - 3). 41. (2, 3) • (1, 1). 42. (1, ll 2 . 

43. (4, - 3) • (4, 3). 44. (0, - l) 3 . 45. (1, 0) • (0, 1) • (1, 1). 

«.«>,■>, _ «-{-r y r)'- 

48. (- \ , - ^) 3 . 49. V~=^ + V^2 - V9. 

50. V" 17 ^ - V 17 ^ - 5t 2 . 51. (4 + 3t) 4- (2 - Si). 

52. ( - 2 + Si) - ( - 6 - Si). 53. (8 + W) + (5 + 2*). 

54. (3 - 2%) - (3 + 5i). 55. ( - 3 + 2t') - (5 - 2t). 

56. (6 + 2i) + (3i + V" 17 !). 57. (2i + 3) + (8 - 5 V 77 !). 

58. (6 + 3i) 4- (3 - 5i). 59. (2 4- 5i) (6 - 3i). 60. 2i (5 + 3i). 

61. (5 + i) (6 + 2i). 62. (3 - 3i) (4t +2). 63. (1 + i V2) (5 -f 2i). 

64. (5i - 8) (2t - 4). 65. (5 - 3t) (2 - 4i). 66. (3 - i) + (2 - 5i). 

67. (3i + 4) -*- (1 - i). 68. (2 + i)» -s- (5 - f). 69. (2 - Si) + (5 - 4i). 

70. (3 - V3i) + (4 + fit)*. 

71. (3 - >/32) 2 + (3 + V3i) 2 - (4 - 3t) (i - 6). 

72. (6 - 2i) (1 4- t) (1 - 3t) -s- (4 - fit). 

73. 1 -5- (6 - 5i). 74. (6 - fit) 4- 1. 75. i + (4 - 3i). 

7fi 3 -4i (4 + i) (i - 5) (3 -I- 50 (8 + 6i) 

' D- (2 4- i) (3 - 2t) " - (6 - 5t) (2t - 7) * '° - (4 - 7t) (4 4- 6i) ' 

79. Prove that complex numbers satisfy the associative and commutative laws of 
addition and multiplication and the distributive law. 

80. Prove that if (a + bi) (c + di) = 0, then a + hi = or c + di = 0. 



Sec. 11-4 



Complex Numbers 



195 



11-4. GRAPHICAL REPRESENTATION 

As we have seen, the complex number a + bi determines a definite 
point P in the plane whose rectangular coordinates are x = a and 
y = b. Conversely, to every point P in the plane corresponds a com- 
plex number a + bi for which the values of a and b are the respec- 
tive rectangular coordinates of P. See Fig. 11-1. In this sys- 
tem, the real numbers a + Oi are represented by points on the 
#-axis, which is called the axis of reals. Pure imaginary numbers 
4- bi are represented by points on the 2/-axis, which is called the 
axis of imaginaries. 




Fig. 11-1. 



P 3 (a + c,b + d) 



*»X 




Fig. 11-2. 



**x 



It is more convenient at times to represent the complex number 
a + bi by the vector drawn from the origin to the point P. The 
length of the vector is given by the relationship r = V a " + b 2 , and 
the direction is given by an angle determined from the equations 
a = r cos and b - r sin 0. 

In Fig. 11-2 is indicated the graphical addition of the two com- 
plex numbers a + bi and c + di, which may be represented either 
by the points P x and P 2 or by the vectors OP 1 and OP^ With^the 
completion of the parallelogram 0P x PJP 2 , the sum of OP 1 and OP 2 
can be represented by the diagonal OP 3 . Thus, either P 3 (a + c, 
b + d) or OP 3 represents the sum (a + c) + (6 + d)i. Hence, the 
vector which represents the sum of two complex numbers is the 
sum of the vectors representing the given numbers. 

To subtract c + di from a + bi graphically, we merely add a + bi 
and — c —di. 



Example 11-3. Add the complex numbers 2 + 3i and 6 -f 2t graphically. 

Solution: Let Pi (2, 3) represent the number 2 -f 3i and let P 2 (6, 2) represent 
the number 6 + 2i. Draw OPi and OP 2 in Fig. 11-3, and complete the parallelo- 
gram OP1P3P2. Then P3_represcnts the sum 8 + 5i* of the complex numbers 
2 + 3t and 6 + 2t", and OP 3 represents the sum of the vectors OPi and OP2. 



196 



Complex Numbers 



Sec. 11- 



P 3 (8,5) 




Fig. 11-3. 

The difference of two complex numbers may be obtained in the 
same manner if we apply the relationship 

(a + bi) — (c + di) = (a + bi) + (— c — di). 

Let P and Q represent the numbers a + bi and c + di, respectively, 
in the complex plane, as shown in Fig. 11-4. Then — c —di is repre- 
sented by Q', which is the reflection of Q through the origin. 




Let us recall how the difference of two vectors was explained in 
Section 6-7 arid was represented graphically in Fig. 6-15. If we let 
the vectors OP, OQ^ and OQ' represent a + bi, c + di, and -c -di, 
respectively, then OR is the desired vector and R is the point 
representing the number (a + bi) — (c + di) . 



11-5. TRIGONOMETRIC REPRESENTATION 

Let the complex number a + bi be represented by the radius 
vector drawn from the origin to the point P. Then the distance 
\OP\ = r is called the modulus, or^the absolute value, of the complex 
number ; and the angle 0, which OP makes with the positive #-axis, 
is called an argument, amplitude, or angle of the complex number. 



Sec. 11-5 



Complex Numbers 



197 



/>(l/2,V3/2) 



From Fig. 11-1, it is clear that - = cos and - = sin 0, or 

a = r cos and b = r sin 0. 
Hence, 

a + 6i = r cos + (r sin 0)i = r(cos + i sin 0). 

This last expression is known as the polar form or the trigono- 
metric form of the given complex number, as contrasted with the 
standard or rectangular form a + bi. 

To reduce a given complex number a + bi to the trigonometric 
form r(cos +i sin 0), we find r and by means of the relation- 
ships r = -\/a 2 + b 2 , a = r cos 0, and 6 = r sin 0. We have 

a + 6i = rf- + - i) = r (cos + i sin 0). 

Example 11-4. Represent the complex number 

1 /3 

o + o~ 1# graphically, and change the given notation 

to the trigonometric form. 

Solution: The point P whose rectangular coordinates 
are ( - > -^- J represents the number - -f —^- i. Since 
a = 1/2 and 6 = a/3/2 are both positive, is a first- 
quadrant angle. We see from Fig. 11-5 that r = 4/-r+7 = l. The angle 6 is 
determined from the equations cos = 1/2 and sin = \/3/2. In this case we may 
let 9 = tt/3 or 60°. Hence, ^ + ^ i = 1 (i + ^ i) = cos 60° + i sin 60°. 

Example 11-5. Express the complex number 1 - i in the trigonometric form. 
Solution: Here a = 1 and b = - 1. So r = \/2, and is a fourth-quadrant angle 

We thus have 




**X 



determined from cos = — -p and sin = — 



V2 * V2 

1 - i = V2(-4= - A=<) = v^(cos 315° + t sin 315°). 



EXERCISE 11-2 

In each of problems from 1 to 12, represent the complex number and its con- 
jugate graphically. 

1. 3 + 2i. 2. 8 + 2*. 3. 3 + 4i. 4. 2 - 3t\ 

5. 3 - 5i. 6. 1 - 1. 7. i. 8. 1. 



9. 



10. |(2 + V2i). 



11. |(1 - v®). 



12. 5 + 12i, 



1 98 Complex Numbers Sec. 1 1-5 

In each of the problems from 13 to 24, perform the indicated operations graph- 
ically. Then check the result algebraically. 

13. (7 - 3t) + (- 4 + i). 14. (2 + 3t) - (4 - 50. 

15. (3 - 60 - (4 + 30. 16. (6 - 20 + (6 + 20. 

17. (3 4- 20 - (5 - 0. 18. (3 + 40 - ( - 2 - 40. 

19. (5 + + (1 - 50. 20. (3 - 20 - (5 - 40. 

21. 3 - (1 - 40 - (2 + 0- 22. (V§ + + (1 + a/30 - 7%. 

23. 7 - (4 - 20 - (- 2 + < \/S). 24. (4 + 30 - (2 + 30 - (3 + 20. 

In each of problems from 25 to 36, change the complex number to the trigono- 
metric form and represent it graphically. 
25. 1 + t. 26. - 5. 27. - 3t\ 28. 3 - 3 y/Zi. 

29. hi + VSi). 30. 1(1 - \/30. 31. 5 + 12*. 32. ?• 

33.6^4t_l^. 84. (3-4f)». 35.^- 36. %=£• 

2 — i 3 +% 6 + 5t 2 -M 

11-6. MULTIPLICATION AND DIVISION IN TRIGONOMETRIC FORM 

Let r x (cos a + i sin a) and r 2 (cos J3 + i sin ft) be any two com- 
plex numbers in trigonometric form. Then, their product is given 
by the relationship 

Ti (cos a + i sin a) • T2 (cos ft + i sin j8) 

= fir2[(cos ol cos ft — sin a sin ft) + t(sin a cos ft + cos a sin ft)] 

= r^cos (a+ ft) + isin (a+ ft)]. 

Thus, we have proved that the absolute value of the product of two 
complex numbers is the product of their absolute values, and an 
angle of the product is the sum of their angles. 

The result found for the product of two complex numbers can be 
extended to the product of three or more complex numbers. 

The quotient obtained by dividing the complex number 
fi (cos a + i sin a) by the complex number r 2 (cos ft + i sin ft) is 
given by the relationship 

ri (cos a + i sin a) __ ri(cosa + t'sina) cos /3 — z sin ft 
r2(cos ft + i sin ft) ~~ r2(cos ft + i sin ft) cos ft — i sin ft 

= — [cos (a — ft) + i sin (a - ft)]. 

It follows that the absolute value of the quotient of two complex 
numbers is the quotient of the absolute values, and an angle of the 
quotient is the difference of their angles. 



Sec. 11-7 Complex Numbers 199 

Example 11-6. Find the product of 2(cos 30° + i sin 30°) and 3(cos 120° + 
i sin 120°). 

Solution: By the rule for products in polar form, we have 
2(cos 30° + i sin 30°) • 3(cos 120° + t sin 120°) 

= 2-3 [cop (30° + 120°) + i sin (30° + 120°)] 

= 6 [cos 150° + i sin 150°] = 6 ( - ^~ + % ) - 3 ( - y/l + i) • 

1 f*\ 
Example 11-7. Find the quotient when 1 — i is divided by ~ + -^— t. 

Solution: From Examples 11-4 and 11-5 in Section 11-5, 
1 - % = \/2(cos 315° + i sin 315°), 
and 

Hence, 



^ + ~ i = cos 60° + i sin 60°. 
1 - i _ ,- (cos 315° + i sin 315°) 



1 , V3 . cos 60° + i sin 60° 

= \/2 [cos (315° - 60°) + i sin (315° - 60°)] 

= V2 [cos 255° + i sin 255°] 

= - y/2 (cos 75° + i sin 75°). 
From a table of trigonometric functions, we find that the result is 

- V2 [0.2588 + i (0.9659)]. 
Or, using exact values of cos 75° and sin 75° previously found, we obtain 

- V2 [^- (V3 - 1) + i^~ ( V3 + 1) J = - \ [(V3 - I) + % (V3 + 1)]. 



11-7. DeMOIVRE'S THEOREM 

If we extend the law of multiplication of the preceding section to 
n factors, we have 

[ri(cos 0i + i sin 0i)] [r 2 (cos 02 + i sin #2)] • • • [r„(cos 6 n + i sin n )] 

= nr 2 ■ • • r n [cos(0i + d 2 + • • + n ) + i sin (0! + 6 2 + • • • + ft.)]. 

If how we put ri — r 2 = • • ■ = r n = r and 0i = 2 = • • • = 9 n = 0, it fol- 
lows that 

[r(cos + ^ sin 0)] n = r n (cos nd + i sin n0). 

This result is known as De Moivre's Theorem. 

Although we have derived De Moivre's Theorem only for integral 
values of n, it can be shown to hold for all real values of n, if prop- 
erly interpreted. 



200 Complex Numbers Sec. 11-7 

Example 11-8. Find the value of (1 - i) 4 by Dc Moivre's Theorem. 

Solution: Since 1 — i = y/2 (—7=- j-i) > the polar form of 1 — i is 

\/2 (cos 315° + i sin 315°). Hence, by De Moivre's Theorem, 
(1 - i) 4 = [y/2 (cos 315° + i sin 315 )] 4 

= ( \/2Y [cos (4 • 315°) + i sin (4 • 315°)] 

= 4(cos 1260° + i sin 1260°) 

= 4(cos 180° + i sin 180°) = - 4. 

Example 11-9. Derive formulas for cos 20 and sin 20 by De Moivre's Theorem. 

Solution: By De Moivre's Theorem for n = 2, we have 
cos 26 + i sin 26 = (cos +i sin 0) 2 = cos 2 6 + (2 cos sin 0) t - sin 2 0. 
The two sides are equal only if the corresponding real and imaginary parts are 
equal. Hence, 

cos 26 = cos 2 6 - sin 2 6, 
and 

sin 20 = 2 sin cos 0. 

11-8. ROOTS OF COMPLEX NUMBERS 

Let p(cos <£ + isin(j)) be an nth root of the complex number 
r(cos + i sin 0), where 0° ;g ;g 360°. Then 

[p(cos <t> + i sin <£)] n = r(cos + i sin 0). 
By De Moivre's theorem, this leads to 
(11-13) p n (cos n<t> + i sin n<f>) = r(cos + i sin 0). 

Our problem now is to find all non-negative numbers p and all 
angles <f> for which (11-13) is satisfied. Separating real and 
imaginary parts, we have 

(11-14) p n cos n<t> = r cos 0, p n sin n<£ = r sin 0. 

Squaring and adding, we obtain 

p 2n (cos 2 n<£ + sin 2 n<f>) = r 2 (cos 2 + sin 2 0). 

Therefore, p 2n = r 2 , since cos 2 a + sin 2 a = 1. The absolute value p 
is then given by the equation 

(11-15) p = y/r. 

From (11-14) we then have 

cos n4> = cos 0, sin n<£ — sin 0. 

It is clear from these equations that the angles ruf) and can differ 
only by a multiple of 2tt or 360°. More precisely, 

(11-16) n<f> = + fc • 360°, or = - + ^^! , 
where 'A: is any integer. 



Sec. 11-8 



Complex Numbers 



201 



For fc = 0, 1, 2, ••• , (n-1) in (11-16), we obtain n distinct 
values of the angle, all of which are non-negative and less than 2rr 
or 360°. Corresponding to these angles we obtain n distinct roots 
given by the formula 



(11-17) 



S^[ 



+ fc.36O° f . . + fc.36O°- 

cos 1_ % Sln _ 



For example, for k — 0, we obtain one nth root, called the prin- 

cipal root, with absolute value r 1/n and angle - : for k = 1, we have 

n 

ft -i— *%f\c\ 
a second root, with absolute value r 1/n and angle — — ; and 

so on to k = n — 1. The value k = n 
would yield the same root as k = 0, 
6 + n • 360° 



n 



since 



cos 



= - + 360°, and 

71 e 



(- + 360°) = cos - and 



sin 







in (- + 360°) = sin - • Similarly, 

k — n + 1 yields the same root as 
k — 1, and so on. This means that 
only n distinct roots exist. 

It is interesting to note that the 
points which represent the roots 
are equally spaced on a circle 
whose radius is ^/r and whose cen- 
ter is the origin. This fact is illus- 
trated in the following example 
and Fig. 11-6. 




Fig. 11-6. 



Example 11-10. Find the three cube roots of 8(cos 60° + i sin 60°). 

Solution: The three cube roots are found by evaluating (11-17). Thus, we have 



>^8(cos 60° + i sin 60°) = -^(cos 



60° + k • 360° 



+ i sin 



60° + k • 360° 



3 ' 3 / 

= 2[cos(20° + k • 120°) + i sin (20° + k • 120°)]. 

As just shown, the substitution of k = 0, 1, 2 yields the three required roots. 
Hence, for k = 0, we have 2(cos 20° + i sin 20°) ; for k = 1, the root is 2(cos 140° + 
isin 140°) ; and for k = 2, the root is 2 (cos 260° + i sin 260°). The roots are repre- 
sented by the equally spaced vectors OP i} OP 2 , and OP 3 , terminating on the circle 
whose radius is 2 and making angles of 20°, 140°*, and 260°, respectively, with the 
positive x-axis. 



202 Complex Numbers Sec. 11-8 

EXERCISE 11-3 

In each of the problems from 1 to 18, perform the indicated operations by first 
expressing the complex number in polar form. Express the answer in rectangular 
form. 

1. (1 + (1 - V3i). 2. (- 1 + t) (V§ + i). 

1 -i 



3. (-1 + V3i)(\/3 +i). 



1 + y/Zi 



K i - \/3 c 2 + 2i 



1 + * V3 + 3t 

7 (1 -t)( z l + V&) , 8 (-1 +*)(a/5 

V3 + 1 * * + * 

11 / I ~ V&V 19 1 +* 

"' V 2 / 1Z ' (2 + i) (3 + i) 

i3. „ .\rj. ,. - 14. 1 - v5i 



(1 + i) (3 + 4i) " ( 2 + 3i) (V3 -f i) 

(2 - 5 i) 2 (1 + 3i) 
(3i) 3 
1 /0A 70 



17. (y/2 - V2t) 10 . 18. 



19. [|(1 - V2i)] 100 ' 20. ( - i + i V3t) 

In each of the problems from 22 to 29, find roots as directed and represent them 
graphically. 

22. Find two distinct square roots of 9 (cos 50° + i sin 50°). 

23. Find four distinct fourth roots of 16(cos 36° + i sin 36°). 

24. Find three distinct cube roots of 27(cos 165° + % sin 165°). 

25. Find five distinct fifth roots of - 32. 

26. Find the three cube roots of 1. 

27. Find the four fourth roots of 1. 

28. Find the two square roots of i. 

29. Find the three cube roots of - •= (1 + \/3i). 



Sec. 11-8 Complex Numbers 203 

In each of the problems from 30 to 34, the complex numbers E f I, and Z designate 
voltage, current, and impedance, respectively, and E = IZ. 

30. Compute E when / = 5 + 4i amperes and Z = 30 — Si ohms. 

31. Compute J when E = 110 + 30i volts and Z = 20 - 15i ohms. 

32. Compute Z when / = 4 + 3t" amperes and ^ = 115 volts. 

33. When two impedances Z\ and Z 2 are connected in parallel, the equation 
-=■=■=-+■=- determines an equivalent impedance Z. Compute Z when Z\ 
= 5 + 4i ohms and Z* = 8 - 6i ohms. 

34. If z and £ are conjugate complex numbers, prove that 

|*+s| 2 +|z-z| 2 =4|s| 2 . 



12 



Equations in 
Quadratic Form 



12-1. QUADRATIC EQUATIONS IN ONE UNKNOWN 

This chapter provides an extension of the work on linear equa- 
tions to second-degree, or quadratic, equations. Consider a quad- 
ratic equation in one unknown written in the form 

(12-1) ax 2 + bx + c = (a ^ 0), 

where a, b, and c are given real numbers. This equation is 
called the general quadratic equation in x, and is said to be in 
standard form. 

If b ¥" 0, (12-1) is called a complete quadratic equation; if b = 0, 
it is called a jmre quadratic equation. Thus, 3#~ — a; + 4 — is a 
complete quadratic equation in which a = 3, 6 = — 1, and c — 4; and 
# 2 — 2 = is a pure quadratic with a = 1 and c = —2. 

In Section 12-4 we shall prove that every quadratic equation has 
two and only two solutions or roots. The roots may be equal or 
unequal, and they may be real or complex. Their natures depend 
on the values of a, b, and c. We shall consider the methods in gen- 
eral use for finding these roots, and we shall then apply them as 
well to the solution of equations which are not quadratic in x but 
which can be written as quadratic equations in expressions involv- 
ing the unknown. 

12-2. SOLUTION OF QUADRATIC EQUATIONS BY FACTORING 

If the left side of a quadratic equation in standard form can be 
factored, the solution of the equation depends on the following 
important principle : 

The product of two or more numbers equals zero if and only if 
at least one of the factors is equal to zero. 

That is, 

A • B = if and only if A = or B = 0. 



Sec. 12-2 Equations in Quadratic Form 205 

In practice, we apply this principle by equating to zero each 
linear factor of the left side of the given quadratic equation, and 
solving the resulting linear equations. The following examples will 
illustrate its application. 



Example 12-1. Solve 2x 2 - 7x + 6 = by factoring. 

Solution: To find the values of x which satisfy the equation 2x 2 — 7x + 6 = 0, 
write the left side in the factored form 

(x - 2) (2x - 3) = 0. 
This product equals zero if and only if either 

x - 2 = or 2x - 3 = 0. 
Hence, x = 2 or x = 3/2. Moreover, 2 and 3/2 are solutions, because both 2 and 
3/2 satisfy 2x 2 - 7x + 6 = 0. Thus, 

2(2)2 _ 7(2) +6=8- 14 +6=0, 
and 

2(3/2) 2 - 7(3/2) +6=9/2- 21/2 +6=0. 

Example 12-2. a) Solve the equation 2 sin 2 x — sin x — 1 = for sin x. b) Find 
all non-negative angles x less than 360° which satisfy this equation. 

Solution: a) Factor the given equation to obtain 

(sin x - 1) (2 sin x + 1) = 0. 

Since sin x — 1 = or 2 sin x + 1 = 0, it follows that sin x = 1 or sin x = — 1/2. 
Chech: 

2(1)2 - (l) - l = 2 - 1 - 1 = 0, 
and 



<-!)'- (-!)->=!+!- — 



0. 



6) When sin .r = 1, x = 90°; when sin x = - 1/2, s = 210° or 330°. Therefore, 
j = 90° or 210° or 330°. 

Check: 2 sin 2 90° - sin 90° - 1 = 2 - 1 - 1 = 0, 

2 sin2 210° - sin 210° - 1 = 2( - ^f- ( - ^) - 1 = 0, 

2 sin2 330° - sin 330° - 1 = 2^ - i) 2 - ( - ^) - 1 = 0. 



and 



Note that this equation is a quadratic in which the unknown is a trigonometric 
function of the angle x. We thus have only two values of sin x which satisfy the 
equation. The determination of the angle, however, goes beyond the algebraic 
solution of the quadratic, and it may happen that there are more than two values 
of x which satisfy the equation. For this reason, it is recommended that all solutions 
be checked by substituting in the original equation. 



206 Equations in Quadratic Form Sec. 12-2 

Example 12-3. Solve the equation 3 sec x = 2 cos x - 1 for all non-negative 
values of x less than 2ir radians. 

Solution: Since sec x = > we can write = 2 cos x — 1. We then clear 

cos x cos x 

of fractions, transpose, and factor, to obtain 

2 cos 2 x — cos x — 3 = (2 cos x — 3) (cos £ + 1 ) =0. 

Hence, cos x = 3/2 or cos a; = - 1. When cos x = - 1, x = 7r. There is no 
real number £ for which cos £ = 3/2. 

3 sec t = 2 cos tt - 1, or 3(- 1) = 2(- 1) - 1. 

It should be noted that factoring provides a method of solving any pure 
quadratic equation ax 2 + c = 0. Thus, the equation is equivalent to 



which gives 



x* + £ = L + kT^\ (x - 4/—^) = 0, 



# + 4/ =0 or # — 4/ = 0, 

" a V a 



or # = ± 4/ • This result agrees with that given by writing x 2 = 

f (X tZ 

and then simply extracting square roots of both sides to obtain 

/ — c c 

x = zb 4/ • Note that the roots are real when -^0 and pure imaginary 

V a a 

when - > 0. 
a 

EXERCISE 12-1 

Solve each of the following equations for x or 0. In each of the problems from 9 
to 20, find all non-negative angles less than 360° which satisfy the given equation. 
Check all solutions. 

1. x 2 + 7x = 0. 2. x 2 - lOx +21 =0. 

3. 2x(Ax + 5) = - 3. 4. x 2 + 6.r - 27 = 0. 

5. x 2 - Z = 6. 6. 6x 2 - 4x - 192 = 0. 

7. x 3 + 27 - 3x(x +3) =0. 8. 16z 2 - a 2 + 2a6 - b 2 = 0. 

9. sin 2 - sin = 0. 10. sin = esc 0. 

11. sin + 1 = 2 esc 0. 12. 2 tan 2 + 3 tan - 2 = 0. 

13. sec (sec + 6) = 16. 14. 3 cos 2 - 8 sin = 0. 

15. (cot - l).(cot + 2) = 4. 16. ?^i * = J2 _ _4 # 

v 3 6 esc 2 esc 

17 cot + 4 143 1ft 8 2+3 cos _ 



cot + 6 ~ (cot + 6) 2 3 cos + 3 ^ 3 cos + 4 

3 7 _ 5 

4 cos 2 8 cos 2 



19. 3 sin 2 0=4 sin 3 0-10 sin 0. 20. , 3 - „ - 77-^ - | = 0. 



12-3. COMPLETING THE SQUARE 

The method developed here is based on the fact that we can make 
any binomial of the form x 2 + kx into a perfect square if we add to 



Sec. 1 2-3 Equations in Quadratic Form 207 

it the square of one-half the coefficient of x. To make this clear, 
let us recall from Section 1-18 the formula for a perfect-square 
trinomial. The formula is 

(x + a) 2 = x 2 + 2ax + a 2 . 

Since the coefficient of x in x 2 4- kx is k, the square of one-half of 

/k\ 2 k 2 
the coefficient is I = ) or -j • Adding this to # 2 + &#, we have 

x 2 + kx + -^ = (a; + ~j • 
Thus, the left member is a perfect square, namely, the square of 

x + r 

Applicability of the procedure to a variety of processes, including 
solution of quadratic equations, is illustrated in the following 
examples. 



Example 12-4. Solve x 2 — 2x — 4 = by completing the square. 

Solution: We first transpose the constant term, so that the left side will be of 
the form x 2 + kx. Hence, the equation becomes 

x 2 - 2x = 4. 

Now the quantity ( — l) 2 = 1 is added to the left side to make it a perfect square. 
To obtain an equivalent equation, the same quantity is added to the right side also. 
The result is 

x 2 - 2x + 1 = 5, 
or 

(x - l) 2 = 5. 

Taking square roots of both sides, we have 

x - 1 = ± y/b. 

So the desired solutions are x = 1 + \/5 and x = 1 — \/h. 
Check: 

j (1 + y/l) 2 - 2(1 4- a/5) -4 = 1-1-2 V5 +5-2-2 >/5 -4=0, 
and _ _ _ 

(1 - V5) 2 - 2(1 - V5) - 4 = 1 - 2 V5 + 5 - 2 + 2 y/b - 4 = 0. 



Example 12-5. Solve 2x 2 - 5x + 3 = by completing the square. 

Solution: Transpose the constant term to obtain 

2x 2 - 5x = - 3. 

Since the coefficient of x 2 is not 1, we make it 1 by dividing both sides by 2. Then 
we have 

2 5 3 

* 2* = ~2* 



208 Equations in Quadratic Form Sec. 12-3 

The square of half the coefficient of x is (^ ( ~~ o) ) = Tp ' ^d tm s numDer to 

both sides, thus making the left side a perfect square. The result is 
2 5 , 25 _ 3 25_J_ 
* ~2* + 16~ ~2 + 16~10* 

5 1 

When we take square roots of both sides, we have x — j = = t • Solving for z, 



we obtain 
That is, 

Check: 
and 



5 , 1 
# = « or x = 1. 



2(1) 2 -5(1) +3=2-5+3=0. 



Example 12-6. Reduce .r 2 + 2/ 2 - 4x + 62/ + 4 = to the form (a: - /i) 2 
+ (1/ - A;) 2 = r 2 . 

Solution: The solution of this problem requires that we complete the square of 
the terms containing y as well as the square of those containing x. Hence, for 
convenience, we write the equation in the form 

Or 2 - 4x ) + (y 2 + %y ) = - 4. 
When we complete the squares in the parentheses, the equation becomes 
(x 2 - Ax + 4) + (y 2 + 6// + 9) = - 4 + 4 + 9. 
Thus, the solution is 

(a? - 2) 2 + (y + 3) 2 = 9. 

Example 12-7. Reduce 9x 2 - Ay 2 - I8x - 16?/ - 43 = to the form 
A(x -h) 2 - B(y -k) 2 = C. 

Solution: Write the equation in the form 9(x 2 - 2x ) - A(y 2 +4?/ ) = 43. 
Complete the squares in the parentheses to obtain 

9(x 2 - 2x + 1) - A(y 2 + 4z/ + 4) = 43 + 9 - 16. 
Note that the numbers 1 and 4, which are added within the parentheses to complete 
the squares, must be multiplied by the coefficients 9 and - 4, respectively, to 
determine the numbers that are added to the right side. 
The reduced form is, therefore, 

9(3 - 1)2 _ 4 (j, + 2)2 = 36. 



Example 12-8. Reduce y/Zx 2 + 4a: - 4 to the form y/a{(x - h) 2 - k 2 ). 

Solution: For convenience, work with the quantity 3x 2 + Ax — 4 without the 
radical sign until the final result is obtained. Hence, write 

3z 2 + Ax - 4 = %(x 2 + |r - |) • 



Sec. 12-4 Equations in Quadratic Form 209 

Complete the square of the terms in x and simplify to obtain 

»(-+J-J)-»((-^+S)-S-!)=»(('+i) , -¥)- 



Now, write this result under the radical sign to obtain 4/3M x +^J — 7j") * 

Comparing this with the required form <\/a((x — h) 2 — k 2 ), we see that we may 
choose a = 3, h = - 2/3, and k = d= 4/3. 

EXERCISE 12-2 

In each of problems from 1 to 15, solve the given equation by completing the 
square. In each of the problems from 9 to 15, find all non-negative angles less 
than 300° which satisfy the given equation. Check all solutions. 
1. x 2 - Sx = 20. 2. x 2 + lOx = 40. 3. x 2 - 7x = 30. 

4. z 2 + £ + 1 = 0. 5. x 2 + x + 2 = 0. 6. 6x 2 - 5x - 1 = 0. 

7. x a + s - 5 = 0. 8. 2z 2 = 3x + 9. 9. tan 2 = 2 tan + 1. 

10. cos 2 = -^+3. 11. ^o=^- 

sec; sec + 2 3 

cop 4-9 
12. 1 + tan-' = sec 0+3. 13. —^n~ = 2. 

esc 2 0—1 

14. 3 cot 2 + cot = 3 - 4 cot 0. 15. sec - cos = 2. 

In each of the problems from 1(5 to 25, reduce the equation to the form 
A(x - h) 2 + B(y - k) 2 = C. 

16. x 2 - Ay 2 - 2x + 1 = 0. 17. x 2 + 4i/ 2 - 6z + 16?/ + 21 = 0. 

18. 4x 2 + 9?/ 2 + 32.Z - lSy + 37 = 0. 19. x 2 + 4// 2 - 10-r - 40?/ + 109 = 0. 
20. 9x 2 + 4?y 2 - 8y - 32 = 0. 21. 4* a + 9</ 2 - lfxc - 18/y -11=0. 

22. x 2 - 9?/ 2 - 4x 4- 36;// - 41 = 0. 23. 4x 2 - 9y 2 + 32x -f 36*/ + 64 = 0. 

24. by 2 - 4r 2 + 50*/ -f- 32x + 41 = 0. 25. 3x* - ?/ 2 + 20x - 2y + 11 = 0. 

In each of the following problems, express the quantity inside the radical or 
parentheses in the form a((x — h) 2 ± A; 2 ). 
26. V* 2 - 6x - 40. 27. v/2x» - 16 j + 41. 28. (4// 2 - 20y - 76) 3 ' 2 . 

29. — 1 = • 30. Or 2 - to + 34) 2 ' 3 . 31. (2x 2 + 28.c + 34)" 1 ' 2 . 

V# 2 — 6x — 7 



32. V(9x 2 + 48.x + 23) 3 . 33. (9x 2 + 24x + 25)" 1 ' 3 . 34. (7.r 2 - 14z + 11)~ 3 . 

12-4. SOLUTION OF QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA 

By applying the method of completing the square to the general 
quadratic equation (12-1), we can obtain a formula for the roots,, 
either real or complex, of any quadratic equation whatever. The 
general equation is 
(12-1) ax 2 + bx + c = 0. 

Transpose the constant term c, and obtain 

ax 2 + bx = — c. 



2 1 Equations in Quadratic Form Sec. 1 2-4 

Dividing by a, we have 



2 f o c 

X * + —x = • 

a a 



/l 6\ 2 ft 2 
Add (?:•-) = t~9 to both sides to obtain 
\2 a/ 4a 2 

o . ft , b 2 b 2 c 

a 4a 2 4a 2 a 

which becomes , , % «, , 2 ,, 

Extracting square roots of both sides gives 



■ ft _ =1= \/b 2 - 4ac 
* + 2a ~ 2a 

Solving for x, we have 



— ft ± s/b 2 — 4ac 
* = £ • 

Hence, to solve a quadratic equation, put it into the standard 
form ax 2 + bx + c = 0, and substitute the coefficients a, b, and c in 
the formula just derived to obtain the roots 



(12-2)si= ^ and z 2 = ^ 

That these numbers Xj. and #2 are solutions of the given quad- 
ratic equation is shown by substituting each of them in (12-1). 
The details of this substitution for Xi follow : 



aXl 2 + bxi + c = ^ _ J + ^ Ya j + c 



- / ft 2 ~ 2b Vb 2 - 4oc + (*> 2 - 4ac) \ 
4a 2 / 



= «(- 



b 2 + b y/ tf—lac 
+ _ +c 



_ b 2 - 2ac - b Vb 2 - 4ac , - 6 2 + & V& 2 - 4ac , 
_ 2a" + 2a + C 

Hence, the number # x satisfies the equation a# 2 + &# + c = 0. A 
similar computation shows that x 2 is also a solution of a# 2 + bx + c 
= 0. Consequently, the two expressions given for # in (12-2) are 
actually roots of (12-1). In Section 12-7 we shall study the expres- 
sions in (12-2) further and shall determine when the roots are dis- 
tinct and when they are real. 



Sec. 12-4 Equations in Quadratic Form 21 1 

Example 12-9. Solve 5x 2 - 6# - 8 = by the quadratic formula. 

Solution: Here a = 5, fr = — 6, c = - 8. Substituting these values in the 

formula, we obtain 

~(-6)±V(-6)2-(4)(5)(-8) 
x _ _ 

_ 6 ± VS6 ± 160 6 ± \Zl96 6 db 14 

10 10 10 

tu f 8 + 14 , 6-14 4 

Therefore, Zi = — r^r — = 2 and x 2 = — jg — = - r • 

These values are seen to satisfy the original equation when substituted for x. 

Example 12-10. Solve x 2 - x + 2 = by the formula. 
Solution: Since a = 1, b =■ — 1, c = 2, we have 

_ 1 d= yT^"8 _ 1 db \/^T _ 1 db \/7i m 

X "" 9 "" 9 "" 9 

Hence, 



JCl = -Z an( J ^ = « 



Check: 



( I ±V7i \* 1 +VTi , _ -6+2V?i 1 + \/7j , 
V 2 / 2 + 2 ~ 4 2 + J 

= -6+2\/7i-2-2V7i+8 :=0 
4 
Similarly, the second root may be checked. 

Example 12-11. Solve 2 sin 2 x + 3 cos x - 3 = by the formula, determining 
all non-negative angles x less than 360°. 

Solution: We make use of the identity sin 2 x + cos 2 x = 1 to transform the 
given equation into one involving a single trigonometric function of x> 
Replacing sin 2 x by 1 — cos 2 x, we have 

2(1 - cos 2 x) + 3 cos x - 3 = 0. 
Simplifying, we obtain 

2 cos 2 x — 3 cos x -f 1 = 0. 
Solving for cos x by the formula, we find 

3 ± V9 -8 3±1 

cos a; = ^ = — z — • 

4 4 

Hence, cos x = 1 or 1/2. Therefore, x = 0° or 60° or 300°. 

The solutions may be checked by substitution in the original equation. 

Example 12-12. Solve cos x tan x + sin 2 x = 1 - sin x by the formula, deter- 
mining all non-negative values of x less than 360°. 

sin x 

Solution: Making use of the identitv tan x = > we have 

cos x 

sin x . . _ i 

cos £ . ^ sm 2 x = i — sin 3. 

cos 3 



2 1 2 Equations in Quadratic Form Sec. 1 2-4 

Transposing and simplifying, we get 

sin 2 x + 2 sin x — 1 = 0. 
Solving for sin x by the quadratic formula, we obtain 

- 2 ± 2 V2 - /o 

sin z = ^ — = - 1 ± v 2. 

The value sin x = — 1 + V2 is one solution. However, sin x = — 1 — \/2must 
be excluded, since the sine of an angle cannot be numerically greater than 1. 

Check: For sin x = \/2 — 1, we find that 

V2 - 1 



cos x = \/2V2 — 2 and tan # = - 



V2V2 - 2 
Substituting these values in the original equation, we have 

V2V2-2- v/2 "~ 1 + (V2 - 1)» = 1 - (V2 - 1). 

V2\/2 - 2 

This reduces to 

V2 - 1 + 3 - 2\/2 = 1 - V2 + 1 
or 

2 - V2 = 2 - V2. 

From the table of trigonometric functions, we have x = 24°28' or 1 55°32'. 

EXERCISE 12-3 

Solve each of the following equations for x or by the quadratic formula. In each 
of the problems from 14 to 24, find all non-negative angles less than 360° which 
satisfy the equation. Check all solutions. 

1. s a + 6x - 7 = 0. 2. x 2 - x - 20 = 0. 

3. x 2 + 2z - 5 = 0. 4. .t 2 + 3 + 1 = 0. 

5. x 2 + 2x + 1 = 0. 6. 7.r 2 - Sx - 9 = 0. 

7. 2x 2 + 3.z + 2 = 0. 8. <Xr 2 - 7x - 5 = 0. 

9. (2x - l) 2 - 2(2.c - 1) - 8 = 0. 10. (a 2 - b 2 )x 2 - iabx - (a 2 - b 2 ) = 0. 

1L J5 _J_ =4 . 12. -A- + 8 - 3 



x+2 2x+3 ' &-3 1-3B+1 

15. --^ ~ = 3. 16. esc - 3 = sin 0. 

tan cot 

17. sin 2 - 2 sin = sin + 3. 18. 16 sec 2 - 8 sec + 1 = 0. 

19 5 , _J_ . 3 - 20 cot * + 2 - 2 cot - 1 _ 

2 "^ cos "*" 2 cos 2 2 cot - 3 cot ~ 

21. (sec + 1) (sec + 2) = sec + 3. 22. (sin + 3) (3 - sin 0) = 3(sin + 3). 

23. (esc + 2) 2 + esc = 1. 24. — l^-r + J = 0. 

cos — 1 Z 

12-5. EQUATIONS INVOLVING RADICALS 

Sometimes an equation in which the unknown appears under a 
radical sign can be reduced to a quadratic by raising both sides to 



Sec. 12-5 Equations in Quadratic Form 213 

a power sufficient to remove the radical. The process must be 
repeated until the unknown no longer occurs under a radical. 

The operation of raising both sides of an equation to a power may 
lead to an equation redundant with respect to the original ; that is, 
the final equation may possess roots that are not roots of the orig- 
inal equation. Such roots are called extraneous roots. For this 
reason, every root obtained must be checked by substitution. 



Example 12-13. Solve the equation y/x 2 - 3a; + 4 = 2. 

Solution: Cube both sides to obtain x 2 - Sx + 4 = 8. Transpose, and get 

x 2 - Sx - 4 = 0. 
Factoring and solving for x, we find that 

x = — 1 or x = 4. 
Check: 



and ^ ( ~ 1)2 " 3( ~ 1} +± = < /i + 3 + 4= V8=2, 

^(4)2 Z 3(4) + 4 = j/s = 2. 
Hence, both — 1 and 4 are roots. 



Example 12-14. Solve the equation y/2x - 1 — V# +3 = 1. 

Solution: Transpose one radical to obtain 

y/2x - 1 = 1 + y/x +3. 
When both sides are squared, the result is 

2x - 1 = 1 + 2aA +3 + z + 3. 

Combining like terms, we obtain 

a: - 5 = 2aA + 3. 
Now we square both sides to get 

x 2 - 10* + 25 = 4(a -I- 3). 
Transposing and combining gives 

:r 2 - 14s + 13 = 0. 
By factoring and solving, we find that 

x = 1 or x = 13. 
Check: 



and V2(l) -1- VI +3 = 1-2*1, 

V2(13) - 1 - V13 +3 =5-4 = 1. 
Hence, 13 is a root, but 1 is not. 

EXERCISE 12-4 

Solve each of the following equations. In each case check for extraneous roots 
1. y/x ~ 2 = 4. 2. V3z +4 = 2. 



3. Vz + 5 = 1. 4. V3z -1=7. 

5. y/x 2 - 16 = 2-r 1 '*. 6. V* 2 - 2 = V2.r + 0. 



214 Equations in Quadratic Form Sec. 12—5 

7. x - 3 - y/xT^l = 0. 8. x - 5s 1 ' 2 +6=0. 



9. V2x + 3 = 4 - 3x. 10. \/z - 1 + \/x - 3 = 2. 

11. \/5 - x + \/4x™T~5 = 5. 12. V2z + <\/2x - 4 = 2. 

12-6. EQUATIONS IN QUADRATIC FORM 

Frequently, an equation which is not quadratic in the given 
unknown may be considered as a quadratic in some expression 
involving the unknown. Thus, ar 4 — 3ar 2 + 2 = and 2(x 2 — 2x) 2 
— (x 2 — 2x) — 6 = may be treated as quadratic equations in the 
expressions ar 2 and (x 2 — 2x), respectively. This type of situation 
was met earlier in Examples 12-11 and 12-12. The following 
examples will further illustrate methods used in solving equations 
in quadratic form. 

Example 12-15. Solve the equation x~* - 3x~ 2 +2=0. 

Solution: Let x~ 2 = y, so that the given equation becomes 

y 2 - Sy + 2 = 0. 
Factor, to obtain 

(y - l) (y - 2) = 0. 
Therefore, 

y = l or y = 2, 
or 

ar» =1 or x~ 2 = 2. 
Hence, the solutions are 

x = =fc 1 and a; = ± — 7= • 

V2 

Example 12-16. Solve (x 2 - 2) 2 - 7(.r 2 - 2) + 10 = 0. 

Solution: Let x 2 — 2 = 2/, so that we get 

?/ 2 - ly + 10 = 0. 
Factor, to obtain 

(V - 2) (y - 5) = 0. 
Hence, 

2/ = 2 or y = 5, 
or 

x 2 - 2 = 2 or .x 2 - 2 = 5. 
Then, 

x 2 = 4 or x 2 = 7, 
and the roots are 

& = ± 2 and x = ± y/l. 



Sec. 12-7 Equations in Quadratic Form 215 



Example 12-17. Solve x 2 + x - 2 y/x 2 + x + 3 = 0. 



Solution: Let V^ 2 + x + 3 = ?/. Then we can write 



(z 2 + z + 3) - 2 -\A 2 + z+3-3=0, or y 2 - 2?/ - 3 = 0. 



Therefore, y = - 1 or 3, and y/x 2 + z+3=-lor y/x 2 + x + 3 = 3. 

By definition of the radical, y/a is a non-nogative number. Hence, although 
y/x 2 +z+3=— lis consistent with the original equation, there are no values 
of x which satisfy this equation. 



Consider, then, y/x 2 + x + 3 = 3. This leads to 

x 2 + x + 3 - 9, or x 2 + z - 6 = 0. 
Hence, 

x = 2 or a = — 3. 
Substitution shows that each of these values of x satisfies the original equation. 



EXERCISE 12-5 

Solve each of the following equations. Check all solutions. 

1. x* + x 2 - 12 = 0. 2. 4x~ 4 - llx- 2 -3=0. 

3. {x 2 + 2) 2 + Six 2 + 2) - 4 = 0. 4. x* - 6x 2 + 8 = 0. 

5. z 4 - 13x 2 + 36 = 0. 6. x 4 - 1 = 0. 

7. (3x - 4) 2 + 6(3* - 4) + 13 = 0. 8. (x 2 + 3z) 2 - 14(x 2 + 3x) + 45 = 0. 

9. (x + -Y+ 2(x + -) - 48 = 0. 10. (x 2 - x) 2 - 20(x 2 - x) + 36 = 0. 
12-7. THE DISCRIMINANT 

It will be recalled that the two roots of the general quadratic 
equation ax- + bx 4- c = are 



no on -b + y/b 2 -An~c -b- y /b 2 - lac 
(12-2) «!= ^ and * 2 = 2a 

The expression 6 2 — 4ac, which appears under the radical sign, is 
called the discriminant of the quadratic polynomial ax 2 + bx + c, or 
the discriminant of equation (12-1). 

In what follows we shall assume that a, b, and c are real, and we 
shall make use of the discriminant to determine the character of 
the roots without actually solving the equation. By inspection of 
the solutions x x and x 2 , we reach the following conclusions : 

1. If b 2 — 4ac = 0, each of the two roots x x and x 2 is equal to 

— — , and both roots are thus real. 
2a 



216 Equations in Quadratic Form Sec, 12-7 

2. If 6 2 — 4ac is positive, then y/b 2 — Aac is real, both roots are 
real, and they are distinct. 

3. If b 2 — 4ac is negative, then \/b 2 — 4ac is imaginary, and the 
roots are distinct complex numbers of the form a + fii and a — f3i. 

These results may be summarized as follows : 



Value of Discriminant 


Character of Roots 


>0 
= 
<0 


Real and unequal 

Real and equal 

Unequal conjugate complex numbers 



Furthermore, if a, b, and c are rational, and b 2 — 4ac is a perfect 
rational square, then the roots are rational; otherwise, they are 
irrational. 

The following examples will illustrate how to determine the 
character of the roots. 

Example 12-18. Determine the character of the roots of 

2x 2 + 7x - 15 = 0. 

Solution: Here a = 2, b = 7, c = — 15. Hence, 

62 - 4ac = 49 + 120 = 169 = (13) 2 . 
The discriminant is positive, and so the roots are real and unequal. Since the 
discriminant is a perfect square, the roots are also rational. 

Example 12-19. Determine all values of k for which the roots of the equation 
kx 2 - 2kx +4=0 are equal. 

Solution: The discriminant must equal zero for the equation to have equal roots. 
Hence, b 2 - 4ac = U 2 - 16& = 4k{k - 4) = 0, and so k - or k = 4. When 
k = 0, the equation is not quadratic. Therefore, the roots are equal only when k = 4. 

EXERCISE 12-6 

Determine the character of the roots of each of the following equations by means 
of the discriminant. 

1. x 2 + 3x + 4 = 0. 2. x 2 + &c - 9 = 0. 3. x 2 + 16x - 6 = 0. 

4. 6z 2 - 7x + 3 = 0. 5. z 2 + lOz + 2 = 0. 6. x 2 + 3x + 2 = 0. 

7. 5z 2 + 7x + 2 = 0. 8. 4x 2 - I2x + 9 = 0. 9. x 2 - 2x + 1 = 0. 

10. 4x 2 - Sx + 2 = 0. 11. 2x 2 - re + 3 = 0. 12. 5x 2 + & + 5 = 0. 

13. 4x 2 - I2x - 9 = 0. 14. x 2 4 4x - 18 = 0. 15. x 2 - 4x + 7 = 0. 



Sec. 12-8 Equations in Quadratic Form 217 

12-8. SUM AND PRODUCT OF THE ROOTS 

Adding the two roots of the general quadratic equation 

ax 2 + bx + c = 0, 
we obtain, by using (12-2), 

» - ft + v^ 2 E 4ac -l ~ b z v 62 E 4ac -_?&-_ ^ 

^ + * 2 - 2a + 2a ~~ 2a ~ a ' 

Also, multiplying these roots, we have 

- b + y/b 2 - Aac - b - \/& 2 - 4ac 

XlX2 = 2a 2a 

— & 2 — (b 2 — 4ac) __ 4ac _ c 
"" 4a 2 ~~ 4a 2 ~~ a 

Hence, we have, for the sum and product of the roots, 

(12-3) xi + x 2 = ~ - > 

and 

(12-4) £i£2 = - • 

These formulas are used in various ways, for example, in check- 
ing roots of a quadratic equation, and in forming an equation if its 
roots are known. 

To find the factored form of the quadratic polynomial ax 2 + bx 
4- c, let us make use of the sum and product formulas just found. 

Solving #i + x 2 = for b, and solving X\X 2 = - for c, we have 

a a 

b = — a(x x + x< 2 ) and c = a(x^x 2 ). Hence, by substitution, we have 

ax 2 + bx + c = ax 2 — a(x\ + #2)x + a(zi£2) 
= a(x 2 — (#i + xi)x + (#1X2)) 
= a(x — x{) (x — #2). 

Therefore, if x A and :r 2 are the roots of the quadratic equation 
ax 2 + bx + c = 0, its factored form can be written 

a(x — x{) (x — X2) = 0. 

Since a^O, we may write the equation equivalently as 

(x — #i) (x — #2) = 0. 

This form or its expansion, 

X 2 — (Xi + X2)X + X1X2 = 0, 

may be used in writing an equation whose roots are known. 
Example 12-21 indicates the procedure. 

Example 12-20. Without solving the equation, find the sum and the product 
of the roots of the equation Sx 2 — 5.c +2=0. 

Solution: In this case, a = 3, b = - 5, and c = 2. Then the sum of the roots is 

= - , and the product of the roots is - =77' 

a 3 a 3 



218 



Equations in Quadratic Form 



Sec. 12-8 



Example 12-21. Write a quadratic equation in the form (12-1), given that the 
roots are (1 ± y/3 i). 

Solution: Since x x = 1 + V3 i and x 2 = 1 — s/S i, we have 

and Xl + X ' 2 = (1 + ^ ^ + (1 " ^ ^ = 2j 

Zi.r 2 = (1 + V3 i) (1 - \/3 i) = 1 - 3 i 2 = 4. 
Therefore a suitable equation is x 2 - 2 c + 4 = 0. 
Alternate Solution: Writing the desired equation in factored form, we have 

(a: - (1 + V3 i)) (a; - (1 - V3 i)) = 0. 
Simplifying, we obtain 

(a; - 1 - V3 (a - 1 + V3 i) = 0, 

((* - 1) - V§ ((* - 1) + V3 f) = 0. 

(a; - l) 2 + 3 = 0, 
and, finally, 

x 2 - 2x + 4 = 0. 



or 
Hence, 



EXERCISE 12-7 

In each of the problems from 1 to 9, find the sum and the product of the roots 
of the given equation. 
1. x 2 + 2x - 1 = 0. 2. 3.r 2 - x + 2 = 0. 3. x 2 + 2 = 0. 

x% \ x + | = 0. 6. 5* 2 



4. 6a: 2 - 2.c + 3 = 0. 



2 4 ' 5 

7. 5.r 2 - 6a: - 1 = 0. 8. 5x 2 + 6* + 1 = 0. 9. lOOx 2 

Form an equation with each of the following pairs of roots. 



11. 0, + 2. 
15. 2 ± tV5. 




12. 
16. 



- 1, 1. 

(1 ± iV3). 



19. 0, V3 - V5. 20. a ± W. 



6a; + 1 = 0. 
- 40x + 17 = 0. 

13. 3, 6. 
17. ±i. 

21. V3 =fc V5 i. 



:*i*o) 

'(-6/2o,o) 



**X 



12-9. GRAPHS OF QUADRATIC FUNCTIONS 

To graph any quadratic function of 
the form ax' 2 + bx + c, we set y = a# 2 
4- 6a; + c and construct a table of values 
of y corresponding to assigned values of 
x. The graph is of the type shown in 
Fig. 12-1 and is called a parabola. 

As found by the quadratic formula, 
the two solutions of the general quad- 
ratic equation (12-1) are given by 



(12-2) 
and 



Fig. 12-1. 



xi = 



X 2 = 



-b + Vb 2 - 


- 4ac 


2a 


- b - Vb 2 ■ 


- Aac 



2a 



Sec. 12-9 Equations in Quadratic Form 219 

Since (12-1) states that y = in the equation y = ax 2 + bx + c, the 
solutions Xi and x 2 are ^-intercepts of the curve. 

In Fig. 12-1, let A and C be the ^-intercept points, and let B be 
the mid-point of AC. We note that 

OB = OA + AB, 

Xi — X 2 Xi + £ 2 

OB = x 2 + - T - = —2— = - ^ • 



Therefore, B is the point f — jr- > J 



Now consider the equation y = fc, which represents a straight 
line parallel to the x-axis. This line may or may not intersect the 
parabola, depending on the value of k. Solving the equations y = k 
and y = ax- + bx -f c simultaneously, by elimination of y we obtain 
ax- + bx + c — fc = 0. The roots of this resulting equation are 

(12-5) xi- -A + 



and 



2a 



* 2 ~ 2a 2a 



v^ 


- 4a(f - 


-k) 


2a 


V& 2 ■ 


- Aa(c - 


-k) 



If the value of k is such that y — k intersects the curve in two 
distinct points, the discriminant in (12-5) is greater than zero, 
and the roots will be real and distinct. The point on the line y — k 
with abscissa x = — ^- is then equidistant from the points of inter- 
section, whose abscissas are x 1 and x 2 . 

If, on the other hand, the value of k is such that y — k does not 
intersect the curve, the discriminant in (12-5) is less than zero, 
and we have a pair of conjugate complex roots, In this case, — — 
is the real part of these roots. 

The points with abscissa x = — — and arbitrary ordinates lie 
on a vertical line, called the axis of symmetry, or simply the axis 
of the curve; the curve is said to be symmetric with respect to 
the axis. 

The point of intersection of the axis and the parabola is called 
the vertex of the parabola. If the coefficient a of the second-degree 
term of y = ax 2 + bx + c is positive, the vertex is the loivest point, 
and the curve is said to be concave upward. If a is negative, the 



220 



Equations in Quadratic Form 



Sec. 12-9 



vertex is the highest point, and the curve is said to be concave 
dotvmvard. 

To find the coordinates of the vertex, we solve simultaneously 



the equation x = — — of the axis and the equation y 

Ad 



ax 2 4- bx + c 
of the parabola. The coordinates of the vertex are thus found to be 



(12-6) 



x = — 



2a 



and 



0-^2 2a + C ~ 4a 



The vertex may be characterized in another way. Let us demand 
that a horizontal line y = k intersect the parabola in two coincident 
points. That is, let us insist that the two roots x x and x 2 in (12-5) 
coincide. Then the discriminant in (12-5) is equal to 0, and 
x x = x 2 = — o- • The value of y corresponding to this value of x is 

Ztd, 

the ordinate of the vertex, as found in (12-6) . We say that the line 

b 2 — 4ac 



y = - 



4a 



is tangent to the parabola at the vertex. 



Example 12-22. Graph x 2 - Gx + 4. 

Solution: Let i/ = x 2 — 6x + 4, assign values to #, and compute the corres- 
ponding values of ?/, as in the accompanying table. The graph is shown in Fig. 12-2. 
The coordinates of the vertex are 

b 6 n b 2 



x= - — = - = 3 and y = c - ~ = 4 
2a 2 4a 



9 = - 5. 



C— 1.11) 




+»x 



X 


V 


- 1 


11 





4 


1 


- 1 


2 


-4 


3 


- 5 


4 


-4 


5 


- 1 


6 


4 


7 


11 



(2,-4) 



Hence, the axis is the line x = 3. Since a is 
positive, the vertex (3, — 5) is the lowest point 
on the curve, and the curve is concave upward. 



Sec. 12-10 



Equations in Quadratic Form 



221 



Example 12-23. Graph y = x 2 - 6a; + 9 and y = x 2 - Qx + 14 relative to the 
same coordinate system as was used for the graph of y = x 2 — 6# -f 4. 

Solution: Tables similar to that in Example 12-22 but applying to the first two 
curves are constructed. The three curves are shown in Fig. 12-3. 



X 


V 


= .t 2 - fxc + 9 


- 1 




16 







9 


1 




4 


2 




1 


3 







4 




1 


5 




4 


6 




9 


7 




16 



X 


£' 


= .r 2 - 6x + 14 


- 1 




21 







14 


1 




9 


2 




6 


3 




5 


4 




6 


5 




9 


6 




14 


7 




21 




>x 



Fig. 12-3. 



Curve (A) crosses the x-axis at two points, corresponding to the roots 3 =fc \/5 
of .r 2 — Gx -\- 4 = 0. Curve (B) is tangent to the .r-axis at (3, 0), because both 
roots of x 2 — 6.c + 9 = are equal to 3. Curve (0) does not intersect the a>axis, 
because 1 x 2 — (ir + 14 — has imaginary roots. 

The reader should relate the discriminants of the quadratics to a study of these 
graphs. 

12-10. QUADRATIC EQUATIONS IN TWO UNKNOWNS 

The general equation of the second degree in x and y is 
(12-7) ax 2 + bxy + cy 2 + dx + ey + / = 0, 

where a, b, c, d, e, and / are given real numbers. An equation of 
this form, in which at least one of the coefficients a, b, and c is 
different from zero, is called a quadratic equation in x and y. 

By a solution of such an equation, we mean a pair of real or 
complex numbers which, when substituted for x and y in (12-7), 
will reduce the left side of the equation to zero. Usually there are 
infinitely many pairs of numbers which satisfy the equation. 



222 Equations in Quadratic Form Sec. 12-10 

If c ¥" 0, (12-7) may be solved for y in terms of x by means of 
the quadratic formula. Corresponding to each real value assigned 
to x, we then obtain, in general, two values of y. We then have 
pairs of numbers (x, y) which, if real, may be plotted in a rec- 
tangular-coordinate system. It is shown in analytic geometry that 
the graph so obtained will be one of a class of curves called conic 
sections, inasmuch as they may be obtained as curves of intersec- 
tion of a plane and a right circular cone. At this time we shall 
confine ourselves to merely listing the curves which comprise this 
class and indicating briefly the form of the quadratic that corre- 
sponds to each of the graphs. A more adequate discussion of this 
subject is given in analytic geometry. However, typical examples 
of these curves are given here. 

1. Parabola. When A^O, the equations y = Ax 1 + Bx + C and 
x = Ay 2 4- By + C represent parabolas with vertical and horizontal 
axes of symmetry, respectively. 

2. Circle. When C is positive, the equation x 2 + y 2 — C represents 
a circle whose center is at the origin and whose radius is \A?- 

3a. Ellipse. When the constants are positive, the equation 
Ax 2 + By 2 = C represents a curve called an ellipse. If A = B, the 
ellipse is a circle. 

3b. Point Ellipse. If A and B are positive and C — 0, the equa- 
tion Ax 2 + By 2 = C is satisfied by only one point, namely, the origin. 
The graph is then said to be a point ellipse. 

3c. Imaginary Ellipse. If A and B are positive and C < 0, there 
are no (real) points on the graph, and we say that the equation 
Ax 2 + By 2 = C represents an imaginary ellipse. 

4a. Hyperbola. When A, B, and C are positive, the equations 
Ax 2 — By 2 = C and Ay 2 — Bx 2 — C represent hyperbolas. 

4b. Hyperbola. When C ¥= 0, the equation xy — C represents a 
curve called an equilateral hyperbola. 

5. Pair of Straight Lines. The equation Ax 2 + Bxy + Cy 2 + Dx 
-f Ey + F = represents two straight lines, which may be either 
distinct or coincident, if the left side can be expressed as the prod- 
uct of two real linear factors. 

We use the quantity b 2 — 4ac, which is usually called the charac- 
teristic of the equation ax 2 + bxy + cy 2 + dx + ey + / = 0, to deter- 
mine the nature of the conic corresponding to a particular form of 
the general quadratic equation. In analytic geometry the following 
statements are shown to be true : 

1. If b 2 — 4ac = 0, the conic is a parabola or two real or imagi- 
nary parallel lines. 

2. If b 2 - 4ac < 0, the conic is an ellipse or a point ellipse or an 
imaginary ellipse. 



Sec. 12-10 Equations in Quadratic Form 223 

3. If b 2 — 4ac > 0, the conic is a hyperbola or two intersecting 
lines. 

In Section 12-9 the graph of the parabola y = ax 2 + bx + c was 
discussed. In the following illustrative examples, the procedures 
for graphs of other quadratic equations are considered. 



Example 12-24. Graph x 2 + y 2 = 9. 

Solution: Set y = to obtain the ^-intercepts, which are ± 3 ; and set x = 
to obtain the ^-intercepts, which are ± 3. Solve the equation for y, obtaining 



y = ± V9 - x 2 . 

Then construct a table of other corresponding values pf x and y. To yield real 
values of ?/, the numerical value of x cannot exceed 3. As shown in Fig. 12-4, the 
resulting graph is a circle with center at the origin and radius 3. 



x 



-4 

-3 

-2 

- 1 



1 

2 

3 

4 



y 



imaginary 





±2.24 


±2.83 


±3 


±2.83 


±2.24 





imaginary 




Fig. 12-4. 



Example 12-25. Graph 4z 2 + 92/ 2 = 36. 

Solution: Set y = to obtain the ^-intercepts, which are ± 3; and set x = to 
obtain the ?/-intercepts, which are ± 2. Solve for y and obtain 



y = ± q V9 - x 2 . 

Construct a table and draw the curve, as shown in Fig. 12-5. This illustrates 
an ellipse. 



X 


y 


-3 





-2 


±1.49 


- 1 


±1.89 





±2 


1 


±1.89 


2 


±1.49 


3 







+x 



Fig. 12-5. 

Note that the numerical value of x must be equal to or less than 3 in order to 
yield real values of y. 



224 



Equations in Quadratic Form 



Sec. 12-10 



Example 12-26. Graph 4x 2 - 9y 2 = 36. 

Solution: Setting y = 0, we find that the x-intercepts are ± 3. Setting x = 0, 
however, results in the equation y 2 = - 4. Hence the curve has no ^/-intercepts. 
Solving the given equation for y } we have 

2 



y = ± 3 v^ 2 



9. 



The accompanying table is constructed. 




O 




X 


y 


-6 


±3.5 


-5 


±2.7 


-4 


± 1.8 


-3 





-2 


imaginary 


2 


imaginary 


3 





4 


±1.8 


5 


±2.7 


6 


±3.5 



Fig. 12-6. 

Note that in this case the numerical value of x must be equal to or greater than 3 in 
order to give real values of y. The graph of the given equation is shown in Fig. 12-6. 
This illustrates a hyperbola. 

EXERCISE 12-8 

Identify and graph each of the following. 
1. x 2 + y 2 = 25. 2. 4x 2 + 9x 2 = 36. 3. 4x 2 - 9y 2 = 0. 

4. 4x 2 - 9y 2 = 36. 5. x 2 - 4*/ 2 = 16. 6. x 2 + 9y 2 = 0. 

7. y = x 2 - 3x + 2. 8. x?/ = - 4. 9. 5a; 2 + 9xy = 28?/ 2 . 

10. 2x 2 + y 2 - 4y =4. 11. 5xy = 2x + y. 

12. 3x 2 - 4x?/ + 2?/ 2 - 6x + 3?/ = 7. 13. x 2 + x# - 2y 2 + 3y - 1 = 0. 

14. 4a- 2 - 4xy -f Z/ 2 - 2x + 4?/ - 12 = 0. 
15. xy + y 2 - y - 2x - 2 = 0. 16. x 2 - 3x - 3?/ 2 + 18y = 27. 

17. 2x 2 - xy - 28# 2 = 0. 18. 4x 2 + \.xy - 3?/ 2 + 4x + 10t/ = 3. 

19. 9x 2 _ 24xy + 16^/2 + 3x - 4?/ = 6. 20. 4x 2 + Zxy + \.y 2 - Sx - %y = 24. 



12-11. GRAPHICAL SOLUTIONS OF SYSTEMS OF EQUATIONS INVOLVING 
QUADRATICS 

In Chapter 9 we solved systems of two or more linear equations 
both algebraically and graphically. Frequently, however, simul- 
taneous systems include one or more equations of the second or 
higher degree. We have, therefore, to consider the problem of find- 
ing systems of values of the unknowns x and y that satisfy two equa- 



Sec. 12-11 



Equations in Quadratic Form 



225 



tions, one of which is quadratic and the other of which is linear or 
quadratic. 

We shall begin by illustrating some graphical solutions of several 
types of systems. The graphical method yields only the real solu- 
tions of a system, but it may prove advantageous in suggesting 
solutions and interpreting results. In general, this method yields 
at best only approximate solutions. The graphs should be drawn as 
accurately as possible. 



Example 12-27. Solve graphically the system 

x 2 - Sx + 3y = 0, * 
x - 3y + 6 = 0. 

Solution: Solving each equation for y in terms of x f we have 

8x - x 2 , x + 6 

y = — 3 — and y = — y- • 

Construct tables of values, and draw both graphs, using the same coordinate 
system, as shown in Fig. 12-7. 





8x - x 2 


X 


y - 3 


- 1 


-3 








1 


7/3 


2 


4 


3 


5 


4 


16/3 


5 


5 


6 


4 


7 


7/3 


8 





9 


-3 



6 




y =■ 



x + 6 




+x 



(A): .r2-.8.l: + 3 l r«0 , 
(*>.r-.3/+6-0 

Fig. 12-7. 

The line and the parabola are seen to intersect at the points (1, 7/3) and (6, 4). 
It follows that these points represent common real solutions, possibly only approxi- 
mate, of the system. A check by substitution shows that the real solutions are, 
in fact, x = 1, y = 7/3; and x = 6, y = 4. 

The solution of Example 12-27 suggests a procedure for finding 
graphical solutions of quadratic equations in one unknown. 



226 



Equations in Quadratic Form 



Sec. 12-11 



Example 12-28. Solve the equation x* - x — 2 = graphically. 

Solution: Since x 2 = x + 2, both sides of this equation may be set equal to y. 
Thus, the original equation is replaced by the system 

J y = z 2 , 

I ?/ = x + 2. 



a; 


2/ = x* 








±1 


1 


±2 


4 


±3 


9 


±4 


16 



X 


y =x +2 



-2 


2 





From the accompanying tables of values of x and y, the graphs shown in Fig. 
12-8 are constructed. The graphs show that the line and the parabola intersect at 
the points (— 1, 1) and (2, 4). Since these points are common to both graphs, 
their abscissas must satisfy the equation 

x 2 - x + 2. 
Hence, the required roots of the original equation x 2 — x — 2 = arc x = — 1 
and x = 2. 





*** 



Fig. 12-8. 



(A): j? 2 +4^- 2 -16 
^; 3.r2_2 r 2 «6 

Fig. 12-9. 



Example 12-29. Solve graphically the system 

x 2 + 4y 2 = 16, 
3x 2 - 2?/ 2 = 6. 

Solution: From the first given equation, 

4?/2 = 16 - x 2 , 
and 



From the second equation, 
and 



V = =fc ^ V16 - z 2 . 
2*/ 2 = 3z 2 - 6, 



Sx 2 -6 



Sec. 12-12 Equations in Quadratic Form 227 

The necessary tables are given here, and the graphs are shown in Fig. 12-9. 





I 


X 


y = =b ^ V16 - z 2 





±2 


1 


±2 Vl5 


2 


db V3 


3 


4V7 


4 





- 1 


±iVl5 


-2 


± a/3 


-3 


= ±V7 


-4 






X 


, i /3^2 - 6 





imaginary 


±1 


imaginary 


=fc v/2 





2 


±V3 


3 


± A/T0i5 


4 


=b V21 


-2 


± \/3 


-3 


± V105 


-4 


±\/21 



The ellipse (A) and the hyperbola (B) are seen to intersect in the following four 
distinct points: 

(2, V3), (2, - V3), ( - 2, V3), ( - 2, - V3). 

These values of x and ?/ already appear in the tables used for constructing the 
graphs, and need not be checked by substitution in the original equations. However, 
such checking is usually desirable. 





EXERCISE 12-9 






Solve each of the following systems of equations graphically. 




1. (x - 2y +3=0, 


2. 


U +?/ =4, 


3. 


(x - y + 1 = 0, 


U 2 = 3y. 




1 ?/ 2 = 2s. 




1*2 + ? y2 = 25. 


4. L> = 3x, 


5. 


|* + 2/ = 6, 


6. 


hx + 22/ = 6, 


Ux + y = 6. 




1 .r 2 = ?/. 




\ X y = - 12. 


7. |2.r -2/ =4, 


8. 


fj* - y* = 16, 


9. 


hx + 32/ = 25, 


Ixj/ = 6. 




1 x + 32/ = 4. 




U*2 + ^2 = 25. 


10. Lr* - 2v/ 2 - 4 = 0, 


11. 


| x 2 + 7/2 = 13, 


12. 


|9z 2 +42/2 =36, 


1*2 _ (ty = o. 




1 x 2 = 122/. 




1*2 + y 2 = si. 


13. jO-r 2 + 25s/ 2 =225, 


14. 


L' 2 +?/ 2 = 20, 


15. 


|*2 -f- j/a = 20, 


}z 2 +t/2=4. 


16. 


\y* -x 2 = 12. 
| X + 7/ = 0, 
{*2 + ^2 = 8. 




[4x 2 + 92/ 2 = 100 



12-12. ALGEBRAIC SOLUTIONS OF SYSTEMS INVOLVING QUADRATICS 

As in the case with linear equations discussed in Section 9-1, it 
may happen that in a system of equations involving quadratics 



228 Equations in Quadratic Form Sec. 12-12 

part of the graph of one equation coincides with part of the graph 
of the other. Such a condition gives rise to infinitely many solu- 
tions. Usually, however, there are only a finite number of points 
of intersection of the graphs corresponding to the given equations, 
and the algebraic problem consists of finding the pairs of numbers 
(x, y) which satisfy both equations. We can say in this case that 
two simultaneous equations in x and y, of degrees m and n, respec- 
tively, can have at most mn solutions. Thus, a system of one linear 
and one quadratic equation can have at most two solutions, and a 
system of two quadratics can have at most four solutions. 

When a system consists of two quadratic equations, the algebraic 
solution usually leads to a fourth-degree equation in one of the 
unknowns. Since we have not presented a general method of solving 
a fourth-degree equation, we shall consider here only systems whose 
solutions can be effected by the theory of quadratic equations. The 
methods of procedure in some of the more important types are 
shown in the following three cases : 

Case 1. One Linear and One Quadratic Equation. A system of 
this type can always be solved by the method of elimination by 
substitution. 

Example 12-30. Solve the system 

x - 3?/ + 6 = 0, 
x 2 - Sx -f 3y = 0. 

Solution: Solve the linear equation for y in terms of x, obtaining 

x + 6 

»=-3— 

Substitution for y in the quadratic equation yields 

x* - 8x + 3 p-jp*) = 0. 

Collecting terms gives 

x 2 - 7x + 6 = 0. 
The roots of this equation are x = 1 and x = 6. Substituting these values in the 
linear equation, we obtain y = 7/3 and y = 4. Hence, the solutions are 

x = 1, y = 7/3; and x = 6, y = 4. 
These values can readily be verified as solutions of the given system. Note that 
they correspond to the coordinates of the points of intersection in Fig. 12-7. 

Example 12-31. Solve the system 

x + 2y + 4 = 0, 

x 2 + 4y* - 2x - 3 = 0. 



Sec. 12-12 Equations in Quadratic Form 229 

Solution: Solve the linear equation for 2y y to obtain 

2y = - (z + 4). 
Substitute in the quadratic and collect terms. We then have 

' x 2 + (a + 4) 2 - 2x - 3 = 0, 
or 

2x 2 + 6x + 13 = 0. 
One solution is 

- 3 + t V17 5 + t V17 

* = 2 ' V= 4 

The other solution is _ 

- 3 - i -v/17 5 - f V17 
*= 2 ' ^ = 4 

Since these values are imaginary, the graphs of the two given equations do not 
intersect. 

Case 2. Two Equations of the Form ax 2 + by 2 = c. When the 
system consists of two equations containing only squared terms in 
each unknown, it can be solved for x 2 and y 2 by the methods used 
for linear systems in Section 9-2. 

Example 12-32. Solve the system 

x 2 + 2y 2 = 17, 

2 X 2 _ y 2 = 14. 

Solution: To eliminate y 2 , multiply the second equation by 2 and add the two 
equations, as follows: 

x 2 + 2y 2 = 17 
4x 2 - 2y 2 = 28 

5.r 2 = 45. 
Solving for x, we have 

x = ±3. 
Now substitute 9 for x 2 in the first of the original equations. Then 

2y 2 = 17 - 9 = 8, 
or 

y = ±2. 

Hence, we have the following four solutions: 

(3,2), (3,-2), (-3,2), (-3,-2). 
These may be written (3, =b 2), ( - 3, d= 2). 

Case 3. Two Equations of the Form ax 2 + bxy + cy 2 = d. If the 
system is of this type, the solution is effected by elimination of the 
constant term. The resulting equation is then solved for one 
unknown in terms of the other. This procedure gives us two linear 
equations in x and y which may be combined with either of the 
given quadratic equations to form two systems of the type con- 
sidered in case 1. 



230 Equations in Quadratic Form Sec. 12-12 

Example 12-33. Solve the system 

x 2 - xy + 2y 2 = 1, 
2x 2 - 2xy + Sy 2 = 3. 

Solution: Multiply the first equation by 3 and subtract the second given equation 
from the new equation, as follows: 

Sx 2 - Sxy + 62/ 2 = 3 

2x 2 - 2xy + Sy 2 = 3 

x 2 - xy - 2y 2 = 0. 
Factor, to obtain 

(x + y) (x - 2y) = 0. 
Hence, 

x + y = or x — 2i/ = 0. 
We may combine each of these two equations with the first given equation to 
form the following two systems: 

f x 2 - xy + 2y 2 = 1, 
I x + y = 0; 

and 

fa; 2 -2;i/+ 2?/ 2 = 1, 
\ a; - 2y = 0. 

We then proceed by the method for case 1. 

The solutions of the given system consist of the solutions of these two systems. 
Hence, we have 

x = 1/2, y = - 1/2; x = - 1/2, y = 1/2; 
x = 1, y = 1/2; * = - 1, ?/ = - 1/2. 
Occasionally, another method is effective in connection with systems described 
under case 3. The following example illustrates this method. 



f & + ®y* = 37, 



Example 12-34. Solve the system 

xy = 2. 

Solution: Multiply the second equation by 6, to obtain §xy = 12. Add this to 
the first equation, to obtain 

x 2 + 6xy + 9ij 2 = 49. 
Also subtract Qxy = 12 from the first equation to obtain 

x 2 - Qxy + $y 2 = 25. 
The left side of each of these new equations is a perfect square. We chose the 
multiplier of xy } which is 6 in this case, so as to obtain perfect squares. We now 
have the system 

/ (x + Sy) 2 = 49, 
\ (x - Sy) 2 = 25. 
Hence, 

x + Sy = 7 or x -f Sy = - 7. 
Also, 

x - Sy = 5 or x - Sy = - 5. 
We now solve the following four systems of linear equations: 
Iz + Sy = 7, ix+Sy= 7, fa + 3y = - 7, f * + Sy = - 7, 

\x-Sy = 5; \x - 3y = - 5; \z-3y=5; \s - 3y = - 5. 



Sec. 12-13 Equations in Quadratic Form 231 

An equation in x and y is said to be symmetric in x and y if the 
equation is unchanged when x and y are interchanged. When a 
system consists of two quadratic equations both of which are sym- 
metric in x and y, the substitutions x = u + v, y — u — v will give 
an equivalent system which may in some cases be solved by previ- 
ous methods. 

EXERCISE 12-10 

In each of the problems from 1 to 21, solve the given system of equations 
algebraically. 

1. f x - 2y + 3 = 0, 2. f x + y = 4, 3. f x - y + 1 = 0, 

\x 2 = 3y. \y 2 = 2x. \x 2 + y 2 = 25. 

4. Uj 2 = 3a;, 5. 13a; + 2y = 6, 6. 1 2x - y = 4, 

1 3a: + y = 6. \xy + 12 = 0. \xy = 6. 

7. jx 2 - ?/ 2 = 16, 8. j 8s + 3y = 25, 9. jy - x = 2, 

jx + 3y = 4. 1 4a; 2 + 2/ 2 = 25. 1 x 2 + y 2 - 2x - 4y = 20. 

10. Is 2 - 2y 2 = 4, 11. \x 2 + y 2 - 10, 12. Uz - 3) 2 + (y - l) 2 = 16, 

\x 2 -9y=0. I a; 2 =9i/. I (x - l) 2 + (?/ - l) 2 = 12. 

13. J 9a; 2 + Ay 2 = 36, 14. \dx 2 + 25y 2 = 225, 15. \x 2 + y 2 = 20, 

} x 2 + 2/ 2 = 81. \ x 2 + 2/ 2 = 4. J?/ 2 - a; 2 = 12. 

16. (xy = 2, 17. 1 1 _ 3 18. (x 2 + 3xy = 10, 

\ X 2 - V 2= 3. a; 2/~2' 1^2/ = 3. 



la; + 2/ = 3. 
19. fa; 2 + Sxy = 28, 20. f 3a; 2 + ?/* = 28, 21. fa;?/ + ?/ 2 = 12, 

[xy + 4?/ 2 = 8. [4a; 2 - xy + y 2 = 40. jxy = 2a; 2 - 24. 

22. Complete the solution of the system in Example 12-34. 

In each of the problems from 23 to 26, solve the given system by the method of 
Example 12-34. 

23. fa; 2 + y 2 = 50, 
\xy = 25. 

25. \x 2 + y 2 = 25, 
(xy = 12. 

Solve each of the following symmetric systems. 
27. fa; 2 + y 2 = 4, 

jxy + a; + y = 10. 
29. frca/y + y 2/ x = 5^ 

\x + 2/ = 2. 

12-13. EXPONENTIAL AND LOGARITHMIC EQUATIONS 

An equation in which the unknown occurs in an exponent is 
called an exponential equation. Such an equation is usually solved 
by taking the logarithm of each side and solving the resulting 
equation. When this latter equation is in linear or quadratic form, 
it may be solved by preceding methods. 





24. fa; 2 + 4y 2 = 13, 




\xy = 3. 




26. fa; 2 + y 2 = 144, 




\xy = 56. 


28. 


fa; 2 + y 2 - x - y = 2, 




\xy + 3a; + 3y = 2. 


30. 


|x 2 + i/ 2 = 13, 




\3x 2 + 2xy + 32/ 2 = 42. 



232 Equations in Quadratic Form Sec. 12-13 

Example 12-35. Solve for x: 5* +3 = 625. 

Solution: Write the equation in the form 

5*+ 3 = 5 4 . 
This equation is satisfied if and only if x + 3 = 4, that is, x = 1. 

Example 12-36. Solve the equation 2 3 * +1 = 3 4 *. 

Solution: Taking the logarithm of each side to the base 10, we get 
log (2 3 * +1 ) = log (3 4 *), 
or 

(Sx + 1) log 2 = 4z log 3. 
Therefore, 

log 2 
4 log 3 - 3 log 2 
From Table III, log 2 = 0.3010 and log 3 = 0.4771. Substituting these values, 
we have 

0.3010 

X ~ 4 • 0.4771 - 3 • 0.3010 ' 
or 

O3010 
X ~ 1.0054 ~ °- 2 " 4 ' 

Example 12-37. Solve for x: log (x + 1) = 1 - log (3s + 2). 

Solution: Collecting terms containing logarithms on one side and writing that 

member as a single logarithm, we have 

log (* + 1) + log (3x + 2) = 1, 

or 

log (x + 1) (3s + 2) = 1. 

Writing the members in exponential form, we have 

(x + 1) (3* + 2) = 10* = 10. 

This equation reduces to 

3x 2 + 5x - 8 = 0. 

§ 

Solving for x, we find that x = lorx= ~^* 

o 

g 

Checking, we find that x = 1 satisfies the original equation, whereas x = — - 

o 
gives rise to logarithms of negative numbers. Since negative numbers do not have 

real logarithms, this latter value of x is not to be used. 



Example 12-38. Solve y = log, (x + y/l + z 2 ) for x in terms of y. 
Solution: Write the given equation in exponential form, obtaining 



& = x + \/l + x 2 . 
Transpose and square to remove the radical. The result is 

e 2 * - 2x& -1=0. 
Solving this equation for x, we have 

e*v - 1 



Sec. 12-14 Equations in Quadratic Form 

EXERCISE 12-11 

Solve each of the following equations for the unknown x. 
1. 2* = 64. 



4. 10* = 0.0001. 

7. 5* = 15. 
10. (3*) (2*) = 36. 
13. log* 2 = 0.6932. 
16. 5.03 = (S.Viyii*- 1 ) 

19. log (3x - 5) = 3 - log 7, 

21. x l0 « ( * 3) = 1000. 
23. e 2 * = 4.83. 

26. e* 2+2 *- 2 = 16. 
e x + e~* 



2. 4*+* = 256. 

5. 4* = 24. 

8. 3* = 17. 
11. 5"* = 2.403. 
14. s 3 « l4 « =0.04681. 
17. 3* +2 = 2(5*). 



233 



3. 3 3 * +l = 243. 

6. 2 4 * = 2 3 *-*. 

9. 3(2*) = 6*. 
12. log, 8 = 0.4136. 
15. (1.5)* = 32. 
18. log 2 x = 3. 



29. y = 



24. c ( * 2) = 9.436. 
27. ac + 6c* +1 = 1. 

e* + e~* 



30. y = 



20. log (4x - 1) = 1 - log (6x + 2). 

22. log 2 (x - 1) + log 2 (x + 3) = 3. 
25. 4e*+* = 7. 

31. e 4 * - e 2 * - 10 = 0. 



e* - e-* 
32. c* + 4e~* = 5. 



12-14. GRAPHS OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS 

The graph ofy = log x is shown in Fig. 12-10. By assigning 
values to x, one finds corresponding values of y from Table III. A 
few pairs of values are shown in the accompanying tabulation. 



X 


y 


0.1 


- 1 


0.2 


-0.70 


0.3 


-0.52 


0.5 


-0.30 


1 





2 


0.30 


3 


0.48 


4 


0.60 


10 


1. 




+x 



Fig. 12-10. 



The graph of y = e* may be obtained from a table of exponential 
functions. Here, however, we shall proceed as follows: Take the 
logarithm of each side to the base 10, to obtain 

log y = x log e. 



234 



Equations in Quadratic Form 



Sec. 12-14 



Prepare the accompanying table, and construct the graph in Fig. 
12-11. 




*►* 



X 


loge 


x log e 


y 





0.434 





1.0 


1 


0.434 


0.434 


2.7 


2 


0.434 


0.868 


7.4 


3 


0.434 


1.302 


20.0 


4 


0.434 


1.736 


54.5 • 



Graph each of the following. 

1. y = log 7 x. 

TT- x i l°gio x 

mnt:log7X =__. 

4. y = 10-* 2 . 
7. y=3+4(20. 
10. 2/ = 100e - 06 *. 



EXERCISE 12-12 

2. x = log 7 ?/. 

5. 7/ = 02. 

8. 2/ = e~*. 
11. y = 7 3 *- 1 . 



3. ?/ = 3-. 

6. 2/ = 16(5-). 

9. y = log c x. 

12. y = 3.9 2 *- 3 . 



13 



Theory of Equations 



13-1. INTRODUCTORY REMARKS 

With the ever-increasing importance of mathematics in engi- 
neering and the physical sciences, problems are constantly occur- 
ring that involve the solution of equations. Often these equations 
are of the simple algebraic or trigonometric types which we have 
already learned to solve. There are many other problems, however, 
which require the solution of equations of higher degree than the 
second and of some types of somewhat more complicated tran- 
scendental equations. Equations of the third and fourth degree can 
be solved by methods analogous to those which we used for quad- 
ratic equations. Because of their complexity, however, these 
methods are seldom used. It has been proved that no such proce- 
dures exist for equations of degree higher than the fourth. 

In this chapter we shall consider various properties of poly- 
nomial equations in general. Some of these properties will be of 
considerable use in later studies of mathematics, while others are 
considered here merely for the aid they give us in determining roots 
of equations. 

13-2. SYNTHETIC DIVISION 

A simplification of the ordinary method of long division, called 
synthetic division, will be presented here. This abbreviated method 
not only enables us to quickly find the quotient and remainder when 

a polynomial f(x) = a x n + a x x n - 1 H V a n is divided by a binomial 

of the form x — r, but also affords a simple process for substituting 
values of the variable into a polynomial. 

We shall divide 2x s — 9x 2 + 13x + 5 by x — 3 to illustrate the pro- 
cedure in synthetic division as compared with that of long division 
considered in Section 1-19. 

235 



236 



Theory of Equations 



Sec. 13-2 



By long division we have : 

2x 3 - 9a; 2 + 13s + 
2a; 3 - 6a; 2 



- 3a; 2 + 13x 
-3a; 2 + 9a; 






4a; + 
4a; — 


5 
12 



x - 3 



2s 2 - 3x + 4 



+ 17 
Thus, the quotient is 2x 2 — Sx + 4, and the remainder is 17. 

Since like powers of x are written in the same vertical column, 
the work may be shortened by writing only the coefficients, as in 
the following schematic arrangement: 



2 


-9 
-6 
-3 


+ 13 
+ 13 


+ 5 


1 -3 




2 


2 -3 


+ 4 










-3 


+ 9 

+ 4 
+ 4 


+ 5 
- 12 







+ 17 
Next, we note that the first term in the divisor x — r need not be 
written, since the divisor is always linear, and the coefficient of x 
in it is always unity. Moreover, it is not necessary to write the first 
term in each row that is to be subtracted, since its coefficient is 
always the same as that of the term directly above. Also, only 
the first term of each partial remainder needs to be written down, 
for the second term is the same as the term directly above it in the 
first row. Finally, the coefficients in the quotient need not be 
written, since these are precisely the leading coefficient in the 
dividend and the remaining partial remainders, excepting the last. 
Hence, we may indicate the process in the following way : 

2-9+13+5 1 -3 
- 6 



-3 



+ 9 



- 12 

+ 17 



Sec. 13-2 Theory of Equations 237 

This scheme can be written compactly as follows : 

2 - 9 13 5 | - 3 

- 6 9 12 
2-3 4 17 

If we replace —3 by +3 in the divisor and add the partial prod- 
ucts in the second row instead of subtracting them, we obtain the 
same result. The synthetic division then takes the following form : 

2-9 13 5 [_3_ 
6 -9 12 



2-3 4 17 



Here the numbers 2, —3, and 4 in the third row are the coefficients 
of the quotient, and the last number 17 is the remainder. 

We can now outline the procedure for synthetic division. Note 
that in every step of the procedure, immediate reference is made 
to the illustrative example. 

To divide f(x) — a x n + a x x n ~ x H h a n by x — r, first arrange 

f(x) in descending powers of x, writing zero for the coefficient of 
any missing power of x. Then arrange the numbers involved in the 
process in three rows, as shown in the following steps : 

Step 1. In the first row, write the coefficients in f(x) in order, as 
a , a U " a n . At the right, put the constant term of the divisor with 
its sign changed. We have then 

Example 
0o ai c&2 • • • a„ | r_ 2 — 9 13 5 | 3_ 

Step 2. Bring down the first coefficient a of f(x) into the first 
place of the third row. Thus, we have 

Example 

ao a\ a,2 • • • a n \_r_ / 2 — 9 13 5 [_3_ 



( 



*a *2 

Step 3, Multiply a by r, and write the product a r in the second, 
row under a x . Bring down the sum of a x and a r into the third row. 
Thus, we now have 

Example 
oo ai fl2 • •• a n |_r_ 2 — 9 13 5 1 3 

apr 6 . 

ao (<W + a>i) 2-3 



238 Theory of Equations Sec. 13-2 

In the example, multiply 2 by 3 and write the product 6 in the 
second row below —9. Then add —9 and 6, writing the sum —3 in 
the third row. 

Step 4. Multiply a r + a x by r, place the product in the second 
row under a 2 , and add. Continue this process until finally a product 
has been added to a n . 

The complete solution of the illustrative example follows : 

2-9 13 5 1 3 

6-9 12 
2-3 4 17 

In the last operations, -3 is multiplied by 3, and the product -9 
is written below 13. The sum of 13 and —9 equals 4. Finally the 
product of 4 and 3, or 12, is written below 5 and added to a n = 5 to 
give 17. 

Step 5. When the process is completed, the last number in the 
third row directly below a n is the remainder. The other numbers 
in this row, read from left to right, are the coefficients of powers of 
x in the quotient arranged in descending order. 

The entire synthetic process of dividing a polynomial f(x) by 
x — r, although it is somewhat complex notationally, can be con- 
veniently exhibited as follows : 

ao ai a,2 • • • a n _i a n | r 

bpr bir • • • b n _ 2 r b n -\r 
bo b\ 62 • • • &, t -i It 

Here the expressions for the coefficients b , b u b 2 , ■ • • , b n 1 of the 
powers of x in the quotient are 6 = a , &i = a„ r + a l9 b< 2 — a r 2 

+ ai r + a 2 • • • , 6 n -i = do ^ n " 1 + Q>i r* 1 ' 2 H + a n -i. Hence, the quotient 

may be written 

(13-1) q (x) = b x*' 1 + 61 x n ~ 2 + ••• + & n -i. 

Also, the remainder assumes the form 

(13-2) R = a r n + air"- 1 H h a n ^r + a n . 

The expression for R is precisely the result of substituting r for x 

inf(x). In other words 

(13-3) «=/(r). 

Finally, if we subtract the product of a? — r and <?(#) from /(#), 
we obtain the remainder R = f(r). Or, if we transpose the product, 
we have the usual statement found in discussions of division. That is, 

(13-4) /(*) = (*-r).a(aO+/(r). 



Sec. 13-2 Theory of Equations 239 

The following examples will further illustrate the process of 
synthetic division. 

Example 13-1. Divide 3x 4 - 4a; 2 + x - 2 by z + 2. 

Solution: Sincere - r = z -f 2, we haver = - 2. Writing zero for the coefficient 
of the missing power x 3 , we have the following result: 

3 0-4 1-2 1-2 

- 6 12-16 30 
3-6 8-15 28 

Note that the first coefficient 3 is brought down into the first place of the third 
row. Next 3 is multiplied by — 2, and the product, which is — 6, is written in 
the second row under 0. The sum of and - 6, or - 6, is written in the third row 
directly belowO. Proceeding in this way, we find that the quotient is 3a? 3 — 6a; 2 + 8a; — 15 
and the remainder is 28. 

Example 13-2. Given /(a;) = a; 3 - 2x 2 + 5s - 4, find 
a)/(-D;&)/(D;c)/(3). 

Solution: By (13-3), f(r) equals the remainder obtained in the division of f(x) 
by a; — r. Hence, we have the following results : 

1-2 5-41-1 



a) 


- 12. 




- 1 


3 


- 8 


Therefore, /( - 1) = - 
b) 


1 
1 


-3 

-2 

1 


8 

5 

- 1 


- 12 

- 4 u 

4 


Therefore, /(l) = 0. 
c) 




1 
1 


- 1 

-2 
3 


4 

5 
3 




- 4 ul 

24 


Therefore, /(3) = 20. 




1 


1 


8 


20 








EXERCISE 


13-1 



In each of the problems from 1 to 15, divide the first function by the second, to 
find the quotient and the remainder, by using synthetic division. 

1. x 2 - Sx + 7, x - 1. 2. a: 3 - x 2 - Sx + 6, x - 3. 

3. x 2 - 5x + 6, x - 4. , 4. a; 3 - 3a; 2 + 6a: - 6, x - 3. 

5. a; 4 - 3a; 2 - 6x + 6, i + 3. 6. a; 3 - 3x 2 + 6.c - 24, a - 4. 

7. 2a; 4 - a; 3 - 6a; 2 + 4a; - 8, a; - 2. 8. 2a; 3 + 3a; 2 + 4, a; -f 2. 

9. .c 3 + 3a; 2 - 2a; - 5, x - 2. 10. 2a; 5 - 3a; 3 + 2a; + 1, x + 2. 

11. a; 4 + x 3 - 59s 2 - 69a; + 030, z 2 - x - 42. (Hint: Factor, the divisor and 
divide successively by each factor.) 



240 Theory of Equations Sec. 13-2 

12. x* - 3a? 3 + 3a; 2 - 3x + 2, x 2 - 3a; + 2. 

13. x 5 + 2a; 4 - 21a; + 18, x 2 + 2x - 3. 

14. a w — 1, a — 1. 

15. x n — i/", x — ?/. 

For each of the following polynomial functions, find the indicated values by the 
method of synthetic division : 

16. f(x) = 3a; 3 - 7x 2 - 5s + 6. Find/(1) and/(- 4). 

17. f(x) = a; 4 - 2a; 3 + 2a; 2 - 5x + 2. Find/( - 2) and/(0). 

18. fix) = x 4 - 4a; 3 - 4a; 2 + 24a; - 9. Find/(3) and/(9). 

19. fix) = a; 4 - 3a; 3 - 13x 2 + 21x + 18. Find/(1) and/(3). 

20. f{x) = 7a; 4 + 37a; 3 - a; 2 - 14a; + 4. Find/(1) and/(- 5). 

13-3. THE REMAINDER THEOREM 

In Section 13-2, it was shown that the remainder in the division 
of a polynomial by a binomial x — r can be found without actually 
performing the division. Thus, in establishing (13-3), we have 
proved the following theorem. 

Remainder Theorem. If a polynomial fix) is divided by x — r, 
the remainder is the value of f(x) for x = r; that is, the remainder 
is/(r). 

It follows from this theorem that f(x) is exactly divisible by 
x — r if and only if f(r) = 0. Hence, we have proved also the fol- 
lowing theorem. 

Factor Theorem. If f(r) = 0, then x — r is a factor of the poly- 
nomial /(#), and conversely. 

Example 13-3. Is x + 3 a factor of a; 4 - 2a; 3 + 3a; 2 - 5? 

Solution: Here f(x) = a; 4 — 2a; 3 + 3a; 2 — 5. Also, x — r = a; -f 3, and r = — 3. 
According to the factor theorem, /( — 3) must equal zero if a; -f 3 is to be a factor 
of f(x). But /(-3) = (-3) 4 -2 (-3) 3 +3(-3) 2 -5 =81 +54 +27 -5 
= 157. Therefore, since /( — 3) ^ 0, x -f 3 cannot be a factor of a; 4 - 2a; 3 + 3a; 2 — 5. 

Example 13-4. Given fix) = a; 3 - 3a; 2 + 5a; - 6. Show that /(2) = and, 
therefore, that x — 2 is a factor of fix). 

Solution: Since /(x) = x 3 - 3x 2 + 5a; - 6, we have by substitution 

/(2) = (2) 3 - 3 (2) 2 + 5 (2) - 6 = 0. 
From the fact that fix) equals zero when x = 2, it follows from the factor theorem 
that x — 2 is a factor of fix). The student should check this result by synthetic 
division and find that 

fix) = a; 3 - 3a; 2 + 5a; - 6 = (a; - 2) (a; 2 - x + 3). 



Sec. 13-4 Theory of Equations 241 

Example 13-5. Find under what condition (x -f a) is a factor of x* + a n , where 
n is an integer and a j* 0. 

Solution: In this case, /(#) = r» + a n , and /( — a) = ( — a) w -f o n . This sum can 
equal zero only if ( — a) n = — a n , that is, only when n is odd. 

EXERCISE 13-2 

In each of the problems from 1 to IS, determine if the second function is a factor 
of the first. If it is a factor, find another factor. 
1. 2x* - Gx 2 + x + 6, x - 2. 2. x* + 4a; 2 + 5z + 6. x 2 + a + 2. 

3. z 3 + 2z 2 - 3x - 1, x - 1. 4. x 7 - 1, x - 1. 

5. a: 3 - x 2 - \\x + 15, x - 3. 6. z fi + 243, x + 3. 

7. x 8 - 256?/ 16 , x - 2?y 2 . 8. 2a; 3 -f 3x* - 9z - 111, a - 1. 

9. z 4 - 4z 3 - x 2 + 16a; - 12, :r + 1. 

10. x 5 + 5a; 4 + 20a; 3 + 60a; 2 + 120z + 120, x + 1. 

11. 5a; 3 4- ah (5 - 6«6 2 )a; a - 2a 2 6 2 (5 + 3a6 2 )a; + 12a 4 6 6 , & - aft. 

12. 2x l - 3a; 3 - 3a; - 2, & + 2. 

13. a; 5 - 10a; 4 + ISa; 3 - 24.1' + 75, x - 2. 

14. 2a; 4 - 31a; 3 + 21a; 2 - 17a; + 10, a; + 1. 

15. 12.x 4 - 40a; 3 - x 2 + Ilia; - 90, 2a; - 3. 

16. 12a; 3 - 22j 2 - 34a; + GO, 3a; + 5. 

17. 24a; 4 - 122a; 3 + 159a; 2 - 7a; - 00, 3a; - 4. 

18. 24x* - 74a; 4 - 85x« + 311a; 2 - 74a; - 120, 2a; + 1. 

19. Show that the equation a; 4 + 2x B - 9r 2 - 2x +8=0 has the roots 1, 2, - 1, - 4. 

20. Find all roots of the equation 12a; 4 - 40a' 3 - x 2 + Ilia; - 90 = 0, given that 
3/2 is a double root, that is, that (a- — 3/2) 2 is a factor of the left side. 

21. Prove that a — b is a divisor of a n — b n for every positive integral value of n. 

22. Prove that a + b is a divisor of a n — b n if n ir? a positive even integer. 

23. Prove that a + b is a divisor of a n + b n if n is a positive odd integer. 

24. Prove that neither a + b nor a — b is a divisor of a n -f b n if n is a positive 
even integer. 

13-4. THE FUNDAMENTAL THEOREM OF ALGEBRA 

We usually assume that every algebraic equation with real or 
complex coefficients has at least one real or complex root. Although 
the existence of such a root is not to be taken for granted, a proof 
is beyond the scope of this book. Accordingly, we shall accept the 
following theorem as true. 

Fundamental Theorem of Algebra. Let f(x) be a polynomial of 
degree n with complex coefficients. Then the algebraic equation 
f(x) =0 has at least one complex root. 

The first complete and rigorous proof of the fundamental theorem 
was given by Gauss in the beginning of the nineteenth century. 
Since that time, many proofs have appeared, but most require 
knowledge of the theory of complex functions. 



242 Theory of Equations Sec. 13-4 

By a repeated application of the fundamental theorem, it can be 
shown that the number of roots of any polynomial equation with 
real or complex coefficients is equal to the degree of the poly- 
nomial. We shall state the following theorem and give its proof. 

Theorem. If f(x) is a polynomial of degree n, the equation 
f(x) = has exactly n roots. 

Proof. Let f(x) be a polynomial of degree n with real or com- 
plex coefficients. By the fundamental theorem, there is a number r x 
such that /(ri) =0. Then, by the factor theorem, 

/Or) = (x - n) • qi(x), 

where qi(x) is a polynomial of degree n — 1 whose leading coeffi- 
cient is a . 

Likewise, qi(x) = has a root by the fundamental theorem. If 
we denote this root by r 2 , then q x (r 2 ) = 0. Also, 

qi(x) = (x - r 2 ) • ?2(s), 

where <z 2 (#) is of degree n — 2 with leading coefficient a . 

Similarly, # 2 (x) = also has a root. We can continue in this way 
until we come to a polynomial of the first degree with root r n . Then 
(13-5) fix) = a (x - n) (x - r 2 ) • • • (x - r n ), 

where a is the coefficient of # in our final first-degree polynomial. 

Since f(x) equals zero when we substitute for x any one of the n 
numbers r u r 2 , • • • , r n , it follows that the equation f(x) = has at 
least the roots r u r 2 , • • • , r w . 

Moreover, there are no other roots. For, suppose that r is some 
root other than r u • • • , r n . Substitution of r for x in (13-5) yields 

f(r) = a (r - ^i) (r — r 2 )"*(r — r n ). 

The right side of this equation cannot equal zero, since no one of the 
factors can equal zero. Therefore, f(r) ^0, and f(x) =0 has no 
more than the n roots found before. 

It may happen that a certain root appears more than once among 
the numbers r lf r 2 ,'> ?V In that case it will be counted as many 
times as the corresponding equal factors appear in (13-5). If a 
certain factor (x — r) appears m times in (13-5), then r is said to 
be a root of multiplicity m. A root is called a simple root, a double 
root, a triple root, and so on, the proper name depending on how 
many times the same factor appears. Hence, combining the state- 
ments that "f(x) = has at least n roots" and "f(x) = has no 
more than n roots," we can conclude that a polynomial equation of 
the nth degree has exactly n roots, a root of multiplicity m being 
counted as m roots. 



Sec. 13-5 Theory of Equations 243 

It is to be noted that a rigorous proof of this theorem requires the 
use of induction. (See Chapter 16.) 

13-5. PAIRS OF COMPLEX ROOTS OF AN EQUATION 

We wish to remind the student that while an equation of the nth 
degree has n complex roots, the number of real roots may be less 
than the degree of the equation. For example, x 2 + 1 = has no 
real roots. Determination of the number of real roots may be sim- 
plified by use of the following theorem. 

Theorem. If all the coefficients of f(x) — are real numbers, and 
if the complex number a + bi is a root of f(x) — 0, then the conju- 
gate a — bi is also a root. It is understood that a and b are real and 
b¥=0. 

Proof. Let x in f(x) — a x n + a x x 71 ' 1 H \- a n be replaced by 

a + bi. Then we have 

f(a + bi) = a (a + bi) n + a x {a + bi) 71 - 1 H h a n . 

If we expand the powers of a + bi by the binomial theorem and 
simplify the resulting expression, then all terms which contain even 
powers of i will be real, while all terms which contain odd powers 
of i will be pure imaginary. Denote by P the aggregate real part, 
and by Q the aggregate imaginary part. Then we have 

f(a + bi) = P + Qi = 0. 

Hence, in accordance with (11-4), P = Q = 0. 

Now, replace x by a — bi in f(x) - 0. In those terms of f(a — bi) 
in which — bi is raised to an even power, the result will remain the 
same as in f(a + bi) . However, all terms in f(a — bi) in which —bi 
is raised to an odd power will have their signs changed. Hence, 

/(a - bi) = P - Qi. 

But, since we have shown that P — Q = 0, we conclude that 

P - Qi = 0. 

In other words, a — £n is also a root of f(x) = 0. 

Example 13-6. Solve x A - z 3 - 2x 2 -{- 6# - 4 = 0, one root being 1 + i. ; 

Solution: According to the last theorem, both 1 + i and 1 — i are roots of the 
given equation. Using Eqs. (12-3) and (12-4) for the sum and product of two 
roots, we find that these conjugate complex numbers are roots of a; 2 — 2x +2 =0. 
Dividing the original polynomial by this quadratic function, we get the quotient 
x 2 + x — 2. Solution of the equation x 2 -j- x — 2 = yields the remaining desired 
roots, x = — 2 and x = 1. 



244 



Theory of Equations 



Sec. 13-5 



EXERCISE 13-3 



1. Solve x 8 + x 2 . - 2x -f 12 = 0, one root being 1 + \/3i. 

2. Solve a: 3 + 3x 2 + 12x - 16, one root being -2-2 V" 17 ^ 

3. Solve z 4 - 2x3 _ j X 2 + 1&. - 18 = 0, one root being 1 - i. 

4. Solve x 4 + 3.t 3 -f 7z 2 + 6z + 4 = 0, one root being - 1 - s/Si. 

5. Solve x* - 8z 4 + 27z 3 - 46x 2 + 38a; - 12 = 0, one root being 2 - V2i, 
and one root being 1. 

6. Solve # 4 + x 2 + 1 = by considering the equation to be a quadratic equation in x 2 . 

7. Find a real cubic equation, two of whose roots are 2 and 1 -f 2i. 

8. Find a real equation of lowest degree having the roots i and 1 — i. 

13-6. THE GRAPH OF A POLYNOMIAL FOR LARGE VALUES OF x. 

In graphing a polynomial function, it is helpful to know the 
location of points on the curve for numerically large values of x. It 
can be shown that, when x is numerically sufficiently large, the term 
a x n of highest degree is numerically larger than the sum of all the 
other terms combined. Therefore, the sign of this term determines 
the sign of the entire polynomial. 

Let us consider the values of the function f{x) = x' 6 + 5x 2 - Ix 
— 13 as x assumes various values from left to right along the £-axis, 
that is, for increasing values of x. The results may be tabulated 
conveniently as follows : 



X 


X 3 


5a; 2 - 7x - 13 


x s + 5x 2 — 7x — 


13 


- 10 


- 1000 


500 + 70 - 13 = 557 


- 443 




- 8 


- 512 


320 + 56 - 13 = 376 


- 136 




- 6 


- 216 


180 + 42 - 13 = 209 


- 7 










0+ - 13 = - 13 


- 13 




5 


125 


125 - 35 - 13 = 77 


202 




10 


1000 


500 - 70 - 13 = 417 


1417 





When x is negative, but numerically sufficiently large, it is seen 
that f(x) is negative. Thus, when x = - 10, x* = - 1000, while the 
sum of other terms, or the value of 5a; 2 - Ix - 13, is only 557 ; and 
/(-10) = -443. For points far enough to the right of the origin, 
say for x = 10, f(x) is positive. For example, /(10) = 1000 + 417 = 
1417. Hence, the graph of y = f(x) would be located below the 
x-axis on the left but would rise above the #-axis as x gets larger 
and larger. 

As a second illustration, let us consider the function f(x) = 
x 4 + Ix 2 -8x + 10. Here f(x) is positive for large numerical values 
of x, regardless of the sign of x. Hence, in this case the graph 



Sec. 1 3-8 Theory of Equations 245 

would be above the #-axis for large numerical values of x to both 
right and left of the origin. 

The following helpful conclusion can be drawn from the preced- 
ing discussion involving large values of x. 

When x is sufficiently large and positive, f(x) has the same sign 
as the leading coefficient a . When x is negative and sufficiently- 
large numerically, f(x) has the same sign as a when n is even, and 
has the opposite sign when n is odd. 

The following symbols are sometimes found in discussions of the 
values of functions for numerically large values of x. When the 
symbolic statement /(+oo)>0is used, what is meant is that, for 
all sufficiently large positive values of x, f(x) ig positive. Similarly, 
the statement /(+ oo ) < means that for all sufficiently large posi- 
tive values of x, f(x) is negative; the statement /(— oo) > means 
that, for all negative values of x which are sufficiently large numer- 
ically, f(x) is positive; and /(— co)< means that, for all negative 
values of x which are sufficiently large numerically, f(x) is negative. 

Thus, if f(x) =x* + 5x 2 -lx- 13,/(- oo) <0 while /(+ oo) > 0. 
However, if f{x) = x* + lx 2 - Sx + 10, /(- oo)>0and/(+ co)>0. 

13-7. ROOTS BETWEEN a AND b IF f(o) AND f(b) HAVE OPPOSITE SIGNS 

Another helpful theorem relating to the roots of a polynomial 
equation is the following. 

Theorem. If the coefficients of a polynomial f(x) are real, and if 
a and b are real numbers such that f(a) and f(b) have opposite 
signs, then the equation f(x) =0 has at least one real root between 
a and b. 

We shall not give a proof of this statement here, but shall merely 
mention the following geometric considerations. The graph of a 
polynomial is a continuous curve; that is, it has no "breaks." There- 
fore, if the points (a,f(a)) and (&,/(&)) lie on opposite sides of 
the #-axis, the graph apparently has to cross the #-axis at least 
once between these points. 

13-8. RATIONAL ROOTS 

The following theorem is fundamental for the solution of equa- 
tions having integral coefficients. 

Theorem, If the equation 

f(x) - a x n + a x x n ~ l H h a n - 

with integral coefficients has the rational root -, > where c and d 

a 



246 Theory of Equations Sec. 13-8 

are integers having no common factor > 1, then c is a divisor of 
the constant term a n , and d is a divisor of the leading coefficient a . 
Proof. We shall make use of a principle from the theory of num- 
bers: If an integer c divides the product of two integers a and b 
and if b and c have no common divisor other than ±1, then c is 
a divisor of a. 

Let -. be a root of f(x) = 0, where c and d are integers with no 
a 

common divisor other than ±1. Then 

pit s*n— 1 ** 

a °T« + ai d^ + '" + an ~ l d + a " = °- 
Multiplication by d n gives 

aoc n + a\c n ~ l d + • • • + a n ^icd n " 1 + a n d n = 0. 

Since c divides all terms before the final one, c also divides that 
term. If now c is factored into primes, none of these primes is a 
divisor of d, and therefore of d n . Thus each prime divides a n , and so 
c itself divides a n . In a similar fashion, it may be shown that d 
divides a . 

Example 13-7. Find the rational roots of Sx 3 - 17z 2 + 15z + 7 = 0. 

c 
Solution: The possible rational roots are of the form ~z > where c is a divisor of 

the constant 7 and d is a divisor of the coefficient 3. The only possible roots are 

± 1, ± 7, ± 1/3, ± 7/3. 
Using synthetic division to check — 1, we obtain the following result: 

3 - 17 15 7 1- 1 

- 3 20 -35 
3 -20 35 -28 

By the remainder theorem, /( — 1) = — 28. Hence, — 1 is not a root. 

We see, however, that /(0) = 4-7. Therefore, by the theorem in Section 13-7, 
there must be a root between x = — 1 and x = 0. Examining our list of possible 
roots, in the hope that a rational root lies between — 1 and 0, we see that a pos- 
sibility is — 1/3. The check by synthetic division follows: 

3 - 17 15 7 1- 1/3 

-16-7 
3 - 18 21 

Hence, x = — 1/3 is a root. After dividing the given polynomial by x + 1/3, and 
equating the quotient Sx 2 - 18x -f 21 to 0, we obtain the quadratic x 2 - 6x + 7 = 0. 
This equation has the roots 3 ± s/2, which are not rational numbers. Therefore, 
the only rational root is x = - 1/3. 

In this example, as is sometimes the case, it has been possible to find all the roots 
of the given equation. 



Sec. 13-8 Theory of Equations 247 

EXERCISE 13-4 

1. Show that 18x 3 - 33a; 2 + 2x + 5 = has real roots between — 1 and 0, 
between and 1, and between 1 and 2. Find these three roots. 

2. Find the rational roots of 2x 3 - 9x 2 + 3x + 4 = 0. 

3. Prove that x 3 + 2x 2 - 3x - 5 = has at least one positive root. 

4. Prove that x 4 - x 3 + x 2 + x - 3 = has at least one positive root and at 
least one negative root. 

5. Prove the following corollary of the theorem in Section 13-8. If f(x) = 
x n + aix*- 1 + * • * + a n = has integral coefficients and has an integral 
root r, then r is a divisor of a„. 

6. Find the integral roots of x 4 - 1 = 0. 

7. Show that the equation x 2 + x + 1 = has no rational roots. 

8. Solve the equation x 3 - 1 = 0. 

9. Find all the integral roots of x 3 + x 2 + x + 1 = 0. 

5 3 

10. Show that 3 is a root of x 3 - -z x 2 - 2x + - = 0. Why does this not contra- 
dict the theorem in Section 13-8 or that in Problem 5? 

In each of the problems from 11 to 20, find all roots of the given equation. 

11. x 4 - 8x 2 + 16 = 0. 12. 4x 3 - 16x 2 - 9x + 30 = 0. 
13. 3x 3 + x 2 + x - 2 = 0. 14. 2x 3 + 3x 2 - 6x - 9 = 0. 
15. 2x 3 - x 2 + 2x - 1 = 0. 16. x 3 - llx 2 + 37x - 35 = 0. 
17. 5x 3 - 13x 2 + 16x - 6 = 0. 18. x 4 - 8x 3 + 37x 2 - 50x = 0. 
19. 2x 3 - x 2 - 4x + 2 = 0. 20. 3x 3 - 13x 2 + 13x - 3 = 0. 



14 



Inequalities 



14-1. INTRODUCTION 

In previous chapters we have explained some methods of deter- 
mining the roots of an equation. By applying these methods, one 
can find the values of an unknown for which a certain function of 
the unknown equals zero. Often, however, it is necessary to solve an 
inequality, that is, to discover for what values of the unknown a 
certain function is less than or greater than another function. 

The present chapter is concerned primarily with the solution of 
inequalities, and the following discussion is essentially an extension 
of the study of the order relation undertaken in Section 1-8. We, 
therefore, recommend that the student thoroughly review Section 
1-8 before starting the study of the present chapter. Since the solu- 
tion of inequalities often involves the use of absolute values, a thor- 
ough mastery of Sections 1-9 and 1-10 is also a requirement. 

The classification of inequalities corresponds to that of equalities 
or equations. As in the study of equations, there are two kinds of 
inequalities involving unknowns, namely, absolute inequalities and 
conditional inequalities. 

An absolute inequality is an inequality that is satisfied by all 
values of the variable or variables for which the functions appear- 
ing are defined. 

A conditional inequality is one that is true only for certain values 
of the variable or variables. Thus, x 2 + 1 > (where x is real) is 
an absolute inequality, because it is true for every real value of x ; 
but x — 1 > is a conditional inequality, because it is valid only 
when x>l. 

14-2. PROPERTIES OF INEQUALITIES 

The rules for dealing with inequalities are to some extent analo- 
gous to those for equations. In transforming inequalities, we shall 
have occasion to use the following elementary principles which 

248 



Sec. 14-3 Inequalities 249 

follow at once from the fundamental properties proved in Section 
1-8. 

Principle 1. If a < b, then a±c<6±c. 
Here are three illustrations : 

From 12 > 8, it follows that 12 + 3 > 8 + 3. 

From 8 < 12, it follows that 8 - 2 < 12 - 2. 

From 12 > 8, it follows that 12 - 8 > 0. 

n h 

Principle 2. If a < b and e> 0, then ac < be and - < - • 

c c 

Two illustrations are given here : 

From 3 < 5, it follows that 3 • 2 < 5 • 2. 

ft 10 

From 8 < 10, it follows that £ < ~ • 

Principle 3. If a < b and c < 0, then ac > be and - > - • 

c c 

Here is an illustration : 

From 3 < 4, it follows that -3 > -4. 

14-3. SOLUTION OF CONDITIONAL INEQUALITIES 

The process of solution of a conditional inequality consists in 
finding all values of the variable which satisfy the inequality. A 
solution consists of a set of values of the variable, rather than one or 
more isolated values as is usual in the case of a conditional equation. 

The discussion in this section is limited to inequalities involving 
rational functions in only one variable. If this variable is x, the 
inequality can be written in the form fix) > or f(x) < 0. For 
instance, suppose we want to solve the inequality 

x 2 - x > 2. 

We may then obtain the following equivalent inequality: 

/Or) = x 2 - x - 2 > 0. 

It is easily seen that this transformed inequality has the same solu- 
tion set as the original inequality. 

In solving this transformed inequality, we find first the values 
of x, if there are any, for which f(x) changes sign as x increases 
in magnitude. If f(x) is a polynomial, such a change of sign occurs 
when f(x) =0. In the example under consideration, changes of 
sign are obtained only at points where 

f{x) = x 2 - x - 2 = 0. 

Hence, we must find the roots of x 2 — x — 2 = 0. These are —1 and 
2, and they determine on the #-axis three intervals throughout each 
of which f(x) retains the same sign. In other words, to find the 



250 Inequalities Sec. 14-3 

solution of f(x) > 0, we find the interval or intervals within which 
f(x) has the sign indicated in the given inequality. In the example 
under consideration this sign is positive, because f(x) is to be > 0. 
The method is applied in Example 14-1. 

In general, the solution of an inequality is obtained by equating 
the function f(x) to zero and solving the resulting equation. If the 

inequality is of the form ^-~ ^ 0, where p(x) and q(x) are poly- 
nomials, it may be cleared of fractions by multiplication by [q (x) ] 2 . 
Since the square of any non-zero real number is positive, the sense 
of the inequality is not changed by multiplication by this factor. 
This leads to the form f(x) ^ 0, where f(x) is a polynomial. The 
values of x for which f{x) changes sign are called critical values. 
When the critical values are arranged in increasing order, they 
determine on the #-axis intervals, throughout each of which f(x) 
cannot change sign. Consequently, the required solution is repre- 
sented by the set of values of x for which f(x) has the same sign 
as that indicated in the given inequality. 

Note. In general, it can be shown that if a factor of f(x) appears 
to an odd power (that is, if f(x) = has roots of odd multiplicity), 
the function will change sign at values of x for which this factor 
vanishes. If a factor of f(x) appears to an even power (that is, if 
f(x) = has roots of even multiplicity), the function will not 
change sign at values of x for which this factor vanishes. There- 
fore, it is sufficient to test only one value of x in one interval. This 
test gives the sign of f(x) in that interval. The sign of f(x) in each 
of the other intervals can be quickly and easily determined from 
the multiplicity of the critical values. Substituting into f(x) a value 
of x in each interval then provides a check of the solution. 

Example 14-1. Solve the inequality 

x 2 - x > 2. 

Solution: An equivalent inequality is 

f(x) = x 2 - x - 2 > 0. 

From the preceding discussion it follows that the critical values are the roots of 

x 2 - x - 2 = 0. 

These roots are — 1 and 2. 

As shown in Fig. 14-1, the points — 1 and 2 determine on the z-axis the following 

three intervals 

(a) x < - 1; (b) - 1 < x < 2; (c) x > 2. 

i ♦ i *■ ♦ 1 > X 

-2-10123 

v „ / v M > v „ / 



(a) (b) (c) 

Fig. 14-1. 



Sec. 14-3 



Inequalities 



251 



Throughout each of these intervals, f(x) retains the same sign. This condition 
may also be seen from the graph of y = x 2 — x — 2 shown in Fig. 14-2. Here 
we note that in each of these intervals the curve lies either entirely above the 
x-axis or entirely below it. 

The solution of f(x) > can now be found by examining the sign of /(#) in each 
of these intervals. Thus, for a value such as x = — 2 in the interval (a), /( — 2) 
= (- 2) 2 - (- 2) -2=4. Hence, f(x) is positive throughout the interval (a). 
In the graph of y = x 2 - x - 2 shown in Fig. 14-2, the curve lies above the 
x-axis to the left of x = — 1. 

For the value x = in the interval (6), 
we have /(0) = - 2. Hence, f(x) is nega- 
tive throughout this interval, and the graph 
lies below the x-axis. 

For the value x = 3 in the interval (c), 
/(3) = (3) 2 - (3) - 2 = 4. So/(x) is posi- 
tive, and the graph again lies above the 
.r-axis. 

These same results may also be obtained 
by the following much shorter procedure: 
Select a value of x in the interval (b) for 
which /(x) is easily evaluated. Such a value 
isx=O.Since/(0) = -2,/(x) <0 throughout 
this interval. Therefore, f(x) > for inter- 
vals (a) and (c), because the sign of f(x) 
changes at the critical values x = — 1 and 
x =2. 

We see, therefore, that in the intervals (a) and (c), f(x) has the sign indicated by 
the given inequality. Since f(x) must be greater than zero, the solution set of the 
given inequality is described by x < - 1 and x > 2. « 




*~X 



y-x 2 -x-2 



Fig. 14-2. 



Example 14-2. Determine the values of x for which \/x 3 - 2x 2 - Sx is real. 

Solution: We shall solve the equivalent problem 

f(x) = x 3 - 2x 2 -3x^0. 
Solving the equation x 3 — 2x 2 — Sx = 0, we find that the critical values are — 1, 0, 3. 



fa 



-l o 



■4- 



3 



-►X 



Fig. 14-3. 



(d) 



As shown in Fig. 14-3, these critical values determine on the x-axis the following 
four intervals: 

(a) x < ~1; (6) - 1 <s <0; 

(c) < x < 3; (d) x > 3. 

Throughout each interval /(x) has the same sign. 

In this example we may select x = 1 in the interval (c). Since /(l) =(1) 3 — 2(1) 2 
- 3(1) = - 4, /(x) < throughout this interval. 



252 Inequalities Sec. 14-3 

As we proceed into the interval (6), we find that/(x) changes sign when x = 0. 
Therefore, f(x) > in the interval (6). Again /(x) changes sign when x = - 1 
and becomes < in the interval (a). Proceeding to the right from the interval (c) 
into the interval (<2), we find that/Or) changes sign when x == 3 and is > in the 
interval (d). 

Thus, we have the following results: 

in (a), /(a) <0; in (&),/(«) > 0; 

in (c), f(x) < 0; in (d), /(x) > 0. 

Hence, the inequality /(x) > is satisfied in intervals (b) and (d). And, since the 
condition x 3 — 2x 2 — 3x = is also allowed in the original problem, the values 
— 1, 0, 3 are included in the solution. Therefore, the solutions for f(x) 
= x 3 — 2x 2 — 3x £ are — 1 £ x £ and x ^ 3. These are the values of x for 
which the original expression V# 3 — 2x 2 — 3x is real. 

x 3 — 3x 2 

Example 14-3. What values of x satisfy ~- > 0? 

X — z 

Solution: Clear the given inequality of fractions by multiplying by (x — 2) 2 and 
obtain /(x) = {x - 2) (x 3 - 3x 2 ) > 0. Solving the equation x 2 (x - 2) (x - 3) = 0, 
we find that the critical values are 0, 0, 2, 3. 

Hence, we have the following four intervals, as shown in Fig. 14-4: 

(a) x < 0; (6) < x < 2; (c) 2 < x < 3; (d) x> 3. 

Throughout each of these intervals /(x) has the same sign. 



*»X 




Let us initially test/(x) for x = 1 in the interval (6). Since /(l) = (1 — 2) [(l) 3 
- 3(1) 2 ] = ( - 1) ( - 2) = 2, it follows that/(x) > throughout this interval. 

We see that/(x) does not change sign for the critical value x = 0, because x = 
is a double root of /(x) = 0. Hence, /(x) > in the interval (a). 

As we proceed to the right from the interval (b) the function /(x) changes sign for 
each of the critical values x = 2 and x = 3. Hence, f(x) < in the interval (c), 
and/(x) > in the interval (fi). 

Therefore, /(x) is positive in the intervals (a), (6), and (d). That is, the solution 
set of the original inequality is described by x < 2, excluding the value x = 0, 
and x > 3. 

Example 14-4. Solve the inequality x 2 - 2x + 3 > 0. 

Solution: Let f(x) = x 2 — 2x + 3. Since the roots of f(x) = are imaginary, 
there are no critical values. Hence, the graph of y = /(x) lies either entirely above 
the x-axis or entirely below that axis. 

Testing for x = 0, we find that/(0) = 3. This result indicates that the graph lies 
above the x-axis. Consequently, the inequality is" satisfied for all real values of x. 



Sec. 14-3 Inequalities 253 

I2x — 1 1 
5 < 1. 

Solution: By Section 1-10, the inequality | x - 6 | < a is equivalent to 

I 2x — 1 I 
6-a<x<6+a. Hence, — r — < 1, which is equivalent to | 2x — 1 1 < 3, 

I o I 

may be written as follows : 

1 - 3 < 2x < 1 + 3 or - 2 < 2x < 4. 
We thus find that the solution set of the original inequality is described by 

- 1 < x < 2. 

Alternate Solution: We may proceed by solving individually the two inequalities 

2x — 1 2x — \ 

- 1 < — r — and — - — < 1 and determining the common solutions; For 

2x — 1 2x — 1 

- 1 < — 5 — i we have - 3 < 2x - 1, or - 1 < x. For — r — < 1, we have 

2x — 1 < 3, or x < 2. The common solutions satisfy the inequalities — 1 < x and 
x < 2. So we again have - 1 < £ < 2. 



EXERCISE 14-1 

In each of the following problems, solve the given conditional inequality or 
inequalities. 



1. x - 3 < 0. 


2. x + 1 < 0. 


3. x + 5 > 0. 


4. x - 1 £ 0. 


5. 4a; - 16 < 0. 


6. 4x - 16 > 0. 


7. 6x + 3 < 0. 


8. 4x - 8 > 3x - 10. 


9. 4s < - 3x - 7, 


10. 3 - 4x < 2x -f 1. 


11. - 3 < 6x < 3. 


12. | x - 1 | < 3. 


M.0<* + 1 +1<1. 


u. _i<£^ + |<i. 


15. | 2x - 3 | < 4. 


16 -IHI< 2 - 


17. z* < 169. 


18. x 2 > 144. 


19. 2x 2 ^ 32. 


»•!-!<«■ . 


21. x(3x +2) < 1, 


22. 4x 2 + 5x < - 1. 


«3<»-i- 


•24.H<-^-l. 

X 2 X 



25. 3x 2 - 3x < - 4. 26. 3x 2 -f 6x < 9. 

27. (x + 1) (x + 2) (x + 3) < 0. 28. (x - 4) (x + 5) (x - 6) > 0. 

29. (x - 1) (x - 2) (x - 3) > 0. 30. (2x - 1) (x + 2) (3x + 1) ^ 0. 



31. y/x 2 - 25 is real. 32. y/x 2 - 5x + 6 is real. 

33. *&£!>! > 0. 34. 4/^T < 1- 

x + 2 r x + 1 

35. |2L±!| < i. 36. -; £ r <0. 

Ixl x 2 — 2x - 3 . 



254 Inequalities Sec. 14-4 

14-4. ABSOLUTE INEQUALITIES 

To prove the truth of an absolute inequality, one must use the 
known properties of the order relation. When none of these seems 
readily applicable to the given inequality, it may be helpful to 
replace this inequality by an equivalent one which may be more 
easily treated. Repeated replacements may have to be made. 

In carrying out a sequence of replacements, one need not verify 
equivalence at each stage, provided that the final inequality can be 
shown to imply the original one. 

The methods for proving absolute inequalities may be used also 
to prove theorems involving inequalities, as in Example 14-7. 

Example 14-6. Prove that a 2 + b 2 ^ 2ab for all real numbers a and b. 

Solution: The given inequality is equivalent, by Principle 1, Section 14-2, to 

a 2 - 2ab + b 2 £ 0, 
that is, to 

(a - b) 2 £ 0. 

This last inequality is true, because the square of every real number is non- 
negative. Therefore, the original inequality is true also. 



a c 
Example 14-7. If a, b, c, and d are distinct positive real numbers, and if -r < -? > 

prove that l<rT1 < r 

Solution: The inequality t < , , is equivalent, by Principle 2, Section 14-2, to 

ab -f- ad < ab + be. 

This inequality is equivalent, by Principle 1, to 

ad < be. 

a c 
This inequality is true because it is equivalent to the given condition r < -i by 

Principle 2. Therefore, the inequality t < , , is true also. 

Similarly, the inequality , , < -j is equivalent, by Principle 2, to 

ad + ed < be + ed, 

and this inequality, by Principle 1, is equivalent to 

ad < be. 

a c 
This last inequality again is equivalent to the given condition r <~j* Hence, 

a A~ c c 
the inequality , , < -j is true also. Therefore, it follows that the original 

,.,. a a + c c . 
inequalities t < , , < -j are true. 



Sec. 14-4 Inequalities 255 

EXERCISE 14-2 

1. Prove that a 2 + 1 ^ 2a if a is real. Note that a 2 + 1 = 2a only when a = 1. 

2. Prove that ^4^ ^ "^TT if a > and 6 > 0. 

2 a + 6 

3. Prove that a 2 > a if a > 1 ; and that a 2 < a if < a < 1. 

4. Prove that a > a 3 if < a < 1. 

5. Prove that t + - > 2 if a and 6 are positive and a j* b. 

o a 

6. Prove that a 2 +b 2 + c 2 ^ ab +bc + ca. (Hint: From Example 14-6, 
a 2 + b 2 £ 2a&, 6 2 + c 2 ^ 26c, and c 2 + a 2 £ 2ca. Add these inequalities.) 

7. If a and b are two positive real numbers, the quantities — - — > \fab, and —tj- 

are called, respectively, the arithmetic mean (A), the geometric mean (G), and the 
harmonic mean (H) of a and b. Prove that H < G < A, except when a = 6. 
(In this case, we have A = G = //.) 

8. Prove that a 3 + & 3 > 3afr(a — 6), if a and 6 are positive and a > b. 

9. Prove that ab + cd g 1, if a, 6, c, and d are positive, and if a 2 + & 2 = 1 and 

c 2 + d 2 = 1. 

10. Prove that 7— ; — ^ t » according as a $ b> if a, 6, and c are positive. 

o + c b 

11. Prove that a*b + ab 3 < a 4 + 6 4 , if a ^ 6. 

12. If z = a; + yi is a c omplex n umber, the modulus or absolute value of z is 
denoted by | z \ = y/x 2 -f- t/ 2 . Show that | z x + z 2 \ £\zi\ +\z 2 \ and that 
I zi — z 2 1 ^ | 2i | — | 32 | , for all complex numbers Z\ and z 2 . 

13. If Zi, z 2 , and Z3 are complex numbers, prove that | Z\ +z 2 1 ^ | Z\ +Z* | -f |z 2 —z% |. 

14. If z is a complex number, prove that ^ | | 2 | + z\ < 2 | z |. 



15 



Progressions 



15-1. SEQUENCES AND SERIES 

Sequences. An infinite sequence, called more simply a sequence 
or sometimes a progression, is a single-valued function whose 
domain of definition is the set of positive integers. A finite sequence 
is a single-valued function whose domain consists of the integers 
1, 2, - - - , m for some positive integer m. 

In specifying a sequence, it is necessary to give a definite rule of 
correspondence which assigns to each integer n a single definite 
number, or term, of the sequence. This term may be denoted by a n . 
In particular, there is a first term ai corresponding to the integer 1, 
a second term a 2 corresponding to the integer 2, and so on. The 
sequence may be specified by the array of numbers 

(15-1) ai, d2, az, • • • , a n , • • • . 

This sequence is denoted briefly by \a n ). (As usual, throughout 
this discussion, a row of dots • • • stands for numbers assumed to be 
present but not written.) A finite sequence has a "last" term and 
may be designated by a l9 a- 2 • • • , a m or simply by {a w }. 

The nth term, or general term, of a sequence is denoted by a n . 
From the rule specifying the nth term for each n, we obtain the 
first, second, third, and other terms of the sequence by substituting 
for n the values 1, 2, 3, and so on in turn. For example, if a n = 1/n, 
the sequence is 

111 

We should note that {a n } is the symbol for the sequence or func- 
tion as a whole, whereas a n is the symbol for the nth term or value 
of the function corresponding to the integer n. 

256 



Sec. 15-1 Progressions 257 

There are many methods for specifying the function in the defini- 
tion of a sequence. Two of these methods follow : 

Explicit Formula, In one method, the nth term is given in terms 
of n itself by means of an explicit formula. 
Here are a few illustrations. 
If a n = ft, the sequence is 1, 2, 3, • • • . 

T r n ,, .12 3 

If a n = 9 , , > the sequence is - > ■= > 77: > • • • . 
rr + 1 2 5 10 

xr n xi. .,23 

If a n = ~ 7 > the sequence is 1, - > « > • • • . 

Zti — 1 * o o 

Recursion Rule. In another method, one or more of the first sev- 
eral values of a n are given explicitly, and a rule is then given 
whereby a n can be calculated from some or all of its predecessors. 

A few illustrations are given here. 

Let a n+ i = a n 2 + 1 with a\ = 0. Then 

a 2 = ai 2 + 1 = 2 + 1 = 1, 
a s = a 2 2 + 1 = l 2 + 1 = 2, 
a 4 = a 3 2 + 1 = 2 2 + 1 = 5, 



Let a n+ i = (n + l)a n with 01 = 1. Then 

a 2 = 2ai = 2 • 1 = 2, 

a 3 = 3a 2 = 3 • 2 = 6, 
a 4 = 4a 3 = 4 • 6 = 24, 



Note that in this example a n = n! 

Let a n +2 = «n+i + «n with a\ = and a 2 = 1. Then 
a 3 = a 2 + ai = 1 + = 1, 
a 4 = a 3 + a 2 = 1 + 1 = 2, 
a 5 = a 4 + a 3 = 2 + 1 = 3, 



Series. Let {a n } be a given sequence of terms ai, 03, • • • , a*, • • ■ . 
Form a new sequence {s n }, where s n is obtained by adding the first 
n terms of \a n }. The sequence of partial sums is then given by 

$1 = aij S2 = ai + a 2 , • • • , 

Sn = ai + a 2 H (- a n , • • • (n = 1, 2, 3, • • •)• 

The sequence {s n } formed in this way is called the (infinite) series 
based on the given sequence {a n }. 

The series as just defined is usually written in the following 
abbreviated form : 

(15-2) ai + a 2 + • • • +On + • • • . 



258 Progressions Sec. 15-1 

Two illustrations of series follow. 

Illustration 1. Let a n = - and s n = ai + a% H I- 0n. Then the 

n 

partial sums are given by 

Sl = ai = 1, 

«2 = &1 + «2 = 1 + o = o ' 

J_ _1_ i.l.l 11 

S3=al + a2 + a3 = l+2 + 3 ::=: "5' , 
The series may then be written 



1 + i + i H + =• + 



n 2 + w ' 



Illustration 2. Let a n = -o-; — and s n = «i + &2 + • • • + a n . Then the 
partial sums are given by 



The series is 



s ± = a x = - , 

j_ 1.1 2 

52 = ai + a2 = o + g = « > 

_i_ . 1,1,1 3 

«3 =ai + a2 + «3 = 2612 = 4 : 



I + I + i, . .. + _!_ + 

2 ^ 6 ^ 12 ^ n 2 + r* 



Limit of a Sequence. One of the most important questions relat- 
ing to sequences is whether or not a given sequence { a n } has a limit 
as n increases indefinitely. If such a limit <£ exists, the sequence 
is said to be convergent or to converge to the limit <£. Symbolically 
this statement may be expressed as follows : 

lim a n = £. 

n-»oo 

This notation is read, "the limit of a n as n increases indefinitely 
is £." 

In order to help make the concept of limit clear, we shall consider 

the sequence given by a n = - • The terms are 

n 

1 1 1 

a\ = 1, a2 = g > as = ^ > • • • > a n = - > • • • . 

When we examine these terms, we notice that the larger the number 
n is, the smaller the term becomes. In other words, as n increases, 

the closer to zero is the number - • In fact, - can be made as small 

n n 

as we please, if we merely choose n sufficiently large. For example, 



Sec. 15-1 Progressions 259 

- < — ; - for every n larger than 100. We see also that - < 



n 100 * n 1000 

for every n larger than 1000. We may conclude that for an arbi- 
trary real positive number d, we can find a value of n, say any 

integer N ^ -j > such that for all integers n> N, it is true that 
- < d. The limit and the sequence I - \ are therefore related as 
indicated by the following statement. 

The sequence j - > converges to the limit <£ = 0, if to each arbi- 
trary positive number d there corresponds an integer N > such 

that — d < - < + d f or every n > N. 

n 

In general, the limit £ of a sequence may be defined as follows : 

Definition of Limit of Sequence. A sequence {a n } converges to the 
limit «£, if to each arbitrary real number d > 0, there corresponds 
a positive integer N such that £— d < a n < £ + d for every n> N. 
This definition may also be put as follows : 

Definition. A sequence { a n ) converges to the limit <£, if for each 
number d > there exists a positive integer N such that 

| £ — a n | < d for every n > N. 

Convergence of a Series. We shall again consider the infinite 
series 

(15-2) ai + a 2 H h a n -) , 

which is the sequence of partial sums 

«li «2i • • • , «m • • • 

of the sequence {a n }. By the following definition the series is con- 
vergent if the sequence of partial sums is convergent. 

Definition. If the sequence of partial sums of the infinite series 
(15-2) converges to a limit, and if lim S n = S, then the series is 

said to converge to the limit S, and S is called the sum of the infinite 
series. 

The new use of the word "sum" for the value S of an infinite 
series is perhaps unfortunate, for it seems meaningless to talk about 
adding up the terms of an infinite series. Actually, S is not a sum, 
but it is rather the limit of a sequence of partial sums of the series. 

If a series does not converge to a limit as n becomes infinite, we 
say that it is divergent, or that it diverges. 



260 Progressions Sec. 15-1 

EXERCISE 15-1 

1. Given a x = 1 and a n +i = ft + a„. Find the five terms of the sequence {a n }. 

2. Given a x = 4, a 2 = 3, and a„ +2 = 2a„ +l — a„. Find the first six terms of 
the sequence { a n } . 

3. Given a x = 1, a 2 = 2, and a„ +£ = (n + l)a n+ i — na n . Find the first six terms 
of the sequence { a n } . 

4. Show that the sequence a n =■ 3 n satisfies the recursion rule a n+ 2 = cin+i + 6a n . 

5. Show that the sequence a n = (— 2) w also satisfies the recursion relationship 
of Problem 4. 

6. Assuming that A and B are any real numbers whatsoever, show that the 
sequence a n = A3 n + B{ - 2) n satisfies the relationship of Problem 4. 

7. Given a n+ i = 2a n + 1 with a x = 2. Let s n = «i + a2 + • • • + a„. Find 
fli, Q>2, ' - • , ct5 and Si, s 2 , • • • , «s. 

8. Show that if s ft = ai + «2 + • • • + «», then s n+ i — s n = a n +i. 

9. Prove that a n = s» — s»_i for n > 1, given ai = s 1# 

10. If s n = ft 2 , show that a n = 2ft — 1 and therefore that 1 + 3 + 5 + • • • 
+ (2ft - 1) = ft 2 . 

15-2. ARITHMETIC PROGRESSIONS 

An arithmetic progression is a sequence of numbers in which 
each term after the first is obtained from the preceding one by 
adding to it a fixed number; this number is called the common 
difference. 

Note that the common difference may be found by subtracting 
any term of the sequence from the one that follows. Thus, 1, 5/2, 
4, 11/2, is an arithmetic progression with the common difference 
3/2, since 11/2 -4 = 4-5/2 = 5/2-1 = 3/2. Also, 5, 1, -3, -7 is 
an arithmetic progression with the common difference —4, since 
-7- (-3) =--3-1 = i_5 = -4. 

It follows, therefore, that a necessary and sufficient condition 
that three numbers A, B, and C form an arithmetic progression is 
C-B = B-A. 

15-3. THE GENERAL TERM OF AN ARITHMETIC PROGRESSION 

Let a denote the first term of an arithmetic progression, and let d 
denote the common difference. Then, by definition, an arithmetic 
progression with n terms may be written as follows : 

a, a + dj a + 2d, a + 3d, • • • , a + (n — l)d. 

Hence if l n represents the value of the nth term, 

(15-3) l n = a + (n - l)d. 



Sec. 15-4 Progressions 261 

Also, we may write an arithmetic progression of n terms in the 
following manner : 

a, a + d, a + 2d, • • • , l n — 2d, Z n — d, l n . 

We shall be concerned with five quantities in connection with an 
arithmetic progression. These are the first term a, the number of 
terms n, the nth term l n , the difference d, and the sum S n of the n 
terms. 

15-4. SUM OF THE FIRST n TERMS OF AN ARITHMETIC PROGRESSION 

Let S n represent the sum of the first n terms of an arithmetic 
progression. If we write the indicated sum in both direct and 
reverse orders, we have 

S n = a + (a + d) + (a + 2d) + • • • + (l n - 2d) + (l n - d) + l n , 
and 

S n = l n + (k - d) + (I» - 2d) + • • • + (a + 2d) + (a + d) + a. 

Adding the right sides, we have n terms each of which is a + l n . 
Thus, 

2S» = (a + l n ) + (a + W) + • • • + (a + t) + (a + l n ) = n(a + J n ). 
Therefore, 
(15-4) S» = | (a + In). 

If we substitute a + (n - l)d from (15-3) for l n in (15-4), we 
have another useful form for the sum. This is 

(15-5) S» = | [2a + (n - l)dj. 

Example 15-1. Determine which of the following sequences are arithmetic 
progressions: 

a) 3, 7, 10; b) 6, 1, - 4; c) 3x - y, 4z + 2/, 5z + 3y. 

Solution: a) Since the differences 10-7 and 7 - 3 are not equal, the sequence 
3, 7, 10 is not an arithmetic progression. 

b) In the second sequence, -4-1=1-6=- 5. Since these differences are 
equal, the sequence 6, 1, — 4 is an arithmetic progression. 

c) We find that (5x + Zy) - (4x + y) = x + 2y and (4z + #) - (3.r - y) 
= £ +2?/. Since there is a common difference, the given sequence is an arithmetic 
progression. 

Example 15-2. Find the twelfth term, and also the sum of the first 12 terms, 

of the arithmetic progression 4, 7, 10, • • ■ . 



262 Progressions Sec. 15-4 

Solution: We have a = 4, n = 12, and d = 3. Then, by (15-3), 
l 12 = a + (n - l)d = 4 + (12 - 1)3 = 37. 
Also, by (15-4), 

flu = \ (a + I.) = y (4 + 37) = 246. 

Example 15-3. The third term of an arithmetic progression is 3/4, and the 
sixth term is 3/2. Find the twenty-second term. 

Solution: By (15-3), h = a -f 2d and U = a + 5d. Thus, we have 

f a + 2d = 3/4, 

\a + 5d = 3/2. 
By solving these two linear equations, we find that a = 1/4 and d = 1/4. By 
(15-3), I22 = 1/4 + 21(1/4) = 11/2. 

Example 15-4. Find each value of x for which the three quantities Zx — 5, x + 4, 
Sx — 2 form an arithmetic progression. 

Solution: Applying the condition C - B = B — A, we have 

(3a: - 2) - (x + 4) = (x + 4) - (3x - 5). 

o 1 • u- 15 ^ ... .. . . 25 31 37 

Solving, we obtain a* = -j- • The arithmetic progression is -j- > -j- » -j- • 

15-5. ARITHMETIC MEANS 

The terms of an arithmetic progression between any two given 
terms are called arithmetic means between the given terms. 

If we let the given terms be a and l H , any number, say k, of means 
may be inserted between a and l n by using the formula l n = a + 
(n — l)d with n — k + 2. As soon as we have found d, we can 
insert the required means. 

Example 15-5. Insert three arithmetic means between and 1 . 

Solution: Let a = 0, l n = 1, and n = 3 + 2 = 5. Then 1 = + 4d, and d = | • 

Hence, the three means are 7 » x » 7 • 

4 2 4 

If only a single arithmetic mean is to be inserted between two 
given numbers, then the inserted value is called the arithmetic 
mean of the given numbers. Thus, if a, x, b form an arithmetic 
progression, x is called the arithmetic mean of a and 6. Since 
b — x = x — a, 

a + 6 

Thus, the single arithmetic mean of two numbers is equal to one- 
half their sum. 



Sec. 1 5-5 Progressions 263 

EXERCISE 15-2 

In each of the problems from 1 to 9, determine if the given sequence is an arith- 
metic progression. Find the next two terms of the extension of each arithmetic 
progression. 
1. 5, 8, 11, 14. 2. - 1, 7, 13, 19. 3. - 10, - 3, 4, 11. 

4.4,12,19,27. «.-H'l.J- ••t'B'-i'-IB- 

7. a, 6, 26 - a, 36 - 2a. 8. a + 6, a — 6, a — 26, a - 36. 9. — - — > a, — ^— • 

In each of the problems from 10 to 19, find l n and S n for the arithmetic progression. 
10. 2, 8, 14, • • • to 12 terms. 11. 3, 6, 9, • • • to 26 terms. 

12. 22, 18, 14, • • • to 7 terms. 13. 1, 2, 3, • • • to 10 terms. 

14. 2, 4, 6, • • • to 50 terms. 15. 1, 3, 5, • • • to 75 terms. 

16. a, 2a, 3a, • • • to 10 terms. 17. 0.2, 0.5, 0.8, • • • to 20 terms. 

18. 1, 8, 15, • • • to 35 terms. 19. 1, 2, 3, • • • to 100 terms. 

In each of the problems from 20 to 27, three quantities relating to an arithmetic 
progression are given. Find the other two quantities. 

20. a = 5, h = 36, n = 4. 21. a = 10, n = 10, d = 10. 

22. S 2 x = 653, n = 21, a = 6. 23. n = 45, d = | , S45 = 63. 

24. Z21 = 8, n = 21, d = ^ • 25. Z» = 7, & = 52, d = | • 

1 3 IS?*! 

26. a = - i , d = ~ , S n = ^- 27. a = 1, S. = 45, d = 1. 

28. Find the sum of the first 100 even integers. 

29. Find the sum of the first n odd integers. 

30. Insert five arithmetic means between 20 and 30. 

31. Insert ten arithmetic means between 100 and 40. 

32. Insert six arithmetic means between — 3 and — 2. 

33. Find the arithmetic mean of 10 and 56 and that of 4 and 28. 

34. Find the arithmetic mean of 28 and 65 and that of 33 and 78. 

35. Insert k arithmetic means between - and a, where a^0, 

a 

36. A display of cans in a grocery store is in the form of a pyramid whose base is an 
equilateral triangle. If each side of the base contains 20 cans and the number of 
cans decreases by one for each successive row, how many cans are in the display? 

37. A lottery contains tickets numbered consecutively from 1 to 100. Customers 
draw tickets and pay according to the number of the ticket, except that 
tickets numbered above 50 cost just 50 cents each. How much money isj 
collected if all tickets are sold? 

38. Determine z so that x, x - 2, Sx will be an arithmetic progression. 

39. The sum of the first and fourth terms of an arithmetic progression is 20. The 
sum of the third and twelfth terms is 36. Find the sum of the first 15 terms. 

40. Find the sum of all multiples of 5 from 100 to 1,000, inclusive. 



264 Progressions Sec. 1 5-6 

15-6. HARMONIC PROGRESSIONS 

A harmonic progression is a sequence of non-zero numbers whose 
reciprocals form an arithmetic progression. Thus, a, b, c are in 

harmonic progression if - > r > - form an arithmetic progression. 

a b c 

The terms of a harmonic progression between any two given 
terms are harmonic means between the given terms. 

To insert a desired number of harmonic means between two 
numbers, we insert the same number of arithmetic means between 
the reciprocals of the two given numbers and then invert the result- 
ing terms. 

The harmonic mean of two numbers is found in the follow- 
ing manner. If a, x, b form a harmonic progression, then - > - > t 

a Qu o 

form an arithmetic progression, and - is the arithmetic mean of - 
I x a 

and r • Hence, 
b 

1 « b 



Solution of this equation for x yields the harmonic mean 

2ab 
a + b 
Clearly the harmonic mean exists only if a + b ¥= 0. 

Example 15-6. Insert three harmonic means between — 3 and 2. 

Solution: The corresponding arithmetic progression is — ■= t • • • > -z • Here a= — ^ > 

1 11 5 

h = - i and n = 5. Hence, - = — - -f 4d, and d = ~j • 

It follows that the arithmetic progression is — - > — o ' To > o7 ' o ' 

24 
Therefore, the three harmonic means are — 8, 12, -=- • 

EXERCISE 15-3 

In each of the problems from 1 to 6, determine if the given sequence is a harmonic 
progression. Find the next two terms of the extension of each harmonic progression. 

t 1 1 JL 9 111 .111 

*" 3'7' if A 4'8'16' 6u 5 , T0 , 15 # 

*• I'!'!- 5. 16,8,^. «.-2,2,f. 

7. Find the tenth term of the harmonic progression - > = > — ' * " ' • 

8. Find the seventh term of the harmonic progression 6, 3, 2, • • • . 

9. Insert four harmonic means between 1 and 2. 



Sec. 15-8 Progressions 265 

10. Insert three harmonic means between ~ and j • 

11. Insert four harmonic means between 6 and 24. 

12. Find the harmonic mean of 6 and 9. 

13. Find the harmonic mean of 24 and 72. 

2 3 

14. Insert nine harmonic means between - and ~ • 

o Z 

15. If a 2 , b 2 , c 2 form an arithmetic progression, show that b -f- c, a + c, a + b form 
a harmonic progression. 

16. If a, b, c form an arithmetic progression and b, c, d form a harmonic progression, 
show that ad = be. 

17. If x is the harmonic mean of a and b, show that 1 r = — h t • 

x — a x — b a b 

18. If a, 6, c, d form a harmonic progression, show that —i = __ , • 

19. If a, 6, c form a harmonic progression, show that - = r • 

20. If the harmonic mean of a and 6 is equal to their arithmetic mean, show that 
a = b y and conversely. 

15-7. GEOMETRIC PROGRESSION 

A geometric progression is a sequence of numbers in which each 
term after the first is obtained from the preceding one by multi- 
plying it by a fixed number; the multiplier is called the common 
ratio. 

The common ratio may be found by dividing any term by the one 

immediately preceding it. Thus, 1, - > - > ^ is a geometric progression 

in which the common ratio is ?:-:-t : = : t- ; -^ = ^- j -1=o' Also, 

8 4 4 2 2 2 

\/2, 1, —.= i - is a geometric progression in which the common ratio 
y2 2 

is - -*- — = = -7= -*- 1 = 1 -5- V 2 = — =• 
2 V2 \/2 V2 

It follows that a necessary and sufficient condition that three non- 

C B 
zero numbers A, B, and C form a geometric progression is -» = -j • 

15-8. THE GENERAL TERM OF A GEOMETRIC PROGRESSION 

Let a denote the first term of a geometric progression, and let r 
denote the common ratio. Then the progression may be written 
as follows : 

a, ar, ar 2 } ar 3 , • • • , ar n ~ l . 

Hence, if l n represents the value of the nth term, 

(15-6) l n = ar*~ l 



266 Progressions Sec. 15-9 

15-9. SUM OF THE FIRST n TERMS OF A GEOMETRIC PROGRESSION 

Let S n represent the sum of the first n terms of a geometric 
progression. Then 

S n = a + ar + ar 2 + • • • + ar n ~ 2 + ar n ~~ l . 

Multiplying by r, we have 

S n r = ar + ar 2 + ar 3 + • • • + ar 11 " 1 + ar n . 

Subtracting the first of these equations from the second, term by 
term, we have 

S n r — S n = ar n — a. 
Therefore, 
(15-7) 5n = o(^_l) = a(^) ( ^ 1} 

If we multiply both sides of (15-6) by r, we get rl n = ar n . Sub- 
stituting in (15-7), we obtain another useful form for S n . This is 

(15-8) S n = ^^ (r ^ 1). 

Note. If r = 1, these formulas do not apply; in this case, however, 

the geometric progression becomes a + aH haton terms, and 

S n = na. 

Example 15-7. Determine which of the following sequences are geometric 
progressions: 

a) 2,6,18; b) 5,10,30; c) x,j>~- 

Solution: a) The ratios found by dividing each of the second and third terms by 
the preceding one are 3 and 3. Hence, the sequence 2, 6, 18 is a geometric progression. 

6) In this sequence, the ratios of consecutive terms are 2 and 3. Therefore, the 
sequence 5, 10, 30 is not a geometric progression. 

c) The given sequence is a geometric progression in which the common ratio 
is x/y. 

Example 15-8. Find U and Ss in the geometric progression 6, 2/3, 2/27, 
Solution: The common ratio is (2/3) -s- 6 = 1/9. Since a = 6 and n = 5, we have 
l 5 = ar*- 1 = 6(l/9) 4 = ■ 6 



ai 65 6! 

( 1 \ 

„ _ a(l - r») _ 6(1 - (1/9)*) _ 6 \ 1 59,049/ _ 14,762 
6 1 - r 1-1/9 ~~ 1-1/9 2187 

Example 15-9. The fifth term of a geometric progression is 3, and the tenth 
term is — 96. Find the common ratio and the first term. 

Solution: By the formula (15-6) for the nth term of a geometric progression, 
we have 

ar* = 3 and ar g = - 96. 



Sec. 15-10 Progressions 267 

Dividing each side of the second equation by the corresponding member of the 
first equation, we obtain r 6 = — 32. Hence, r = - 2. Therefore, a( - 2) 4 = 3, or 
3 

Example 15-10. Find each value of x for which the three numbers x, x — 2, 

x + 1 form a geometric progression. 

C B x 4- \ x — 2 
Solution: If we apply the condition -5- = -j 1 we have ^ = • Solving, 

Jj A X — m Z X 

we obtain x =4/5. Hence, the geometric progression is 4/5, — 6/5, 9/5. 

15-10. GEOMETRIC MEANS 

Terms of a geometric progression between any two given num- 
bers are called geometric means between the given numbers. Let a 
and l n be given numbers. Then k means may be inserted between 
them by using the formula l n = ar*- 1 with n = k + 2. 

Example 15-11. Insert three geometric means between 1 and 2. 

Solution: Let a = 1, l n = 2, and n = 5. Then Z 5 = cm* 4 and r = 2 1/4 . Hence, 
the three means are 2 1 ' 4 , 2 1 ' 2 , 2 3 ' 4 . 

If a, x, b form a geometric progression, then x is called a greo- 

b x 
metric mean of a and 6. Since - = - » 

a; a 

x = db V^- 

Hence, we note that a geometric mean of two numbers is the same 
as a mean proportional between the two numbers. 

EXERCISE 15-4 

In each of the problems from 1 to 9, determine whether the given sequence is a 
geometric progression. Find the next two terms of the extension of each geometric 
progression. 

1. 2, 8, 32. 2. ! , - 1, 2. 3. 4, 16, 64. 

4. 2,4,6. 5. 27,18,12. G# ^'i' A* 

_ V3 V6 V3 k n a* a\ Q 1 1 1 

In each of the problems from 10 to 15, find l n and £ n in the given geometric 
progression. 

10. 4, 2, 1, • • • ; n = 10.. 11. 3, 2, 4/3, • • • ; n = 15. 

12. 3, 9, 27, • • • ; n = 45. 13. 100, - 10, 1, • • • ; n = 101. 

14. log 2, log 4, log 16, • • • ; n = 10. 15. log 9, log 3, log \/3, • • • ; n = 6. 

16. Find the sum of the first n terms of the geometric progression 1, ^ > j > • • • . 



268 Progressions Sec. 15-10 

17. For what values of x do x - 2, x — 6, 2x + 3 form a geometric progression? 

18. For what values of x do 3x -f 4, a; — 2, 5z -f 1 form a geometric progression? 

19. For what values of x is x -f 1 a geometric mean of 2x + 1 and x — 1? 

20. Find a geometric mean of 4 and 16. Also, find their arithmetic mean. 

21. Find a geometric mean and the arithmetic mean of 3 and 12. 

22. Insert four geometric means between 1 and 32. 

23. Insert five geometric means between 1 and 1,000,000. 

24. Insert ten geometric means between 1 and 2. 

25. If A, G, and H denote the arithmetic mean, a geometric mean, and the harmonic 
mean, respectively, of two numbers a and 6, prove that G 2 = AH. 

26. If the arithmetic mean of a and 6, when a, b ^ 0, is A, a geometric mean is G, 
and the harmonic mean is 77, find A — G,G — H, and A — H. 

2 4 

27. Show that the sum of the first n terms of the geometric progression 1, ^ > ^ > • • • 

is 3(1 — (o))* Discuss how this sum varies as n increases. 

28. Show that the sum of the first n terms of the geometric progression 8, 4, 2, • • • is 
16 (l — (o) ) ' Discuss how this sum varies as n increases. 

29. Find the sum of the first n terms of the sequence 1, 2x } 3x 2 , 4a: 3 , • • • . (Hint: 
Let S n be the sum. Then compute S n — xS n .) 

30. Find the sum of the terms of the finite sequence 2, = > =^ > -=^ > • • • > — =^ — • 

15-11. INFINITE GEOMETRIC PROGRESSION 

A geometric progression in which the number of terms is infinite 
is called an infinite geometric progression. 

In Section 15-9, we found an expression for the nth partial sum 
S n of a geometric progression. Hence, S n is the nth term of the 
geometric series based on the given progression. Thus, we have 

Si = a; S2 = a + ar; S3 = a + ar + ar 2 ; • • • ; S n = a + ar + • • • + ar"- 1 . 

Furthermore, 

„ _ q(l - r») , . 

s n - 1 _ r fr * 1), 

or 

1 — r 1 — r 

Let us consider what happens to the sum of n terms of a geo- 
metric progression when the number of terms increases indefinitely. 

If a = 0, evidently S n = whether r ¥" 1 or r = 1. In this case, the 
number meets the requirement of a limit of S n , so that the sum S 
of the infinite geometric series is equal to 0. 

Now suppose a ¥* 0. For this condition, we consider four cases. 

Case 1. Assume that \r\ < 1. If r ^ 0, the numerical value of r* 
decreases as n increases. Moreover, by making the number of terms 



Sec. 15-12 Progressions 269 

sufficiently large, we can make \r n \ as small as we please. It follows 
that if \r\ < 1, we can make S n differ from by as little as we 

1 — T 

please ; that is, S n approaches -- as a limit. This condition may 
be stated symbolically in the following manner: 

(15-9) 8 = lim S n = a 



1 - r 

where S is the sum of the infinite geometric series. Equation (15-9) 
is true also if r = 0, since in this case S n has the constant value a. 

Case 2. If r = 1, then S n = na. Since a ¥" 0, |S„| increases indefi- 
nitely as n increases. Here { S n J diverges. 

Case 3. If \r\ > 1, then Ir 71 ] increases indefinitely as n increases. 

fl (lT n 

Hence, so does I S n I = I z z I • Again {S n } diverges. 

1 — r 1 — r 

Case I>. If r = — 1, then the progression becomes, a, —a, a, —a, • • • , 
(— l) w x a. If n is even, S n = 0. If n is odd, S n = a. In this case, we 
say that S n oscillates between and a. Here also the series diverges. 

The sum of an infinite geometric progression can therefore be 
found by (15-9), but only when \r\ < 1. When r = 1, r — — 1, or 
|r| > 1, the series diverges, and we say that the series has no sum. 
For a further discussion of this topic, the student may refer to a 
treatise on the theory of limits. 

Example 15-12. The owner of a fleet of trucks finds that if used motor oil is 
refined for re-use, 20 per cent of the oil is lost in the process. If he starts with 100 
gallons of refined oil and re-refines this oil each time it becomes dirty, determine 
the total amount of oil he has used before the entire 100 gallons is lost. 

Solution: We begin with 100 gallons. After the first reclaiming operation, we 
have 80 gallons of good oil. When this becomes used and is re-refined, we have 64 
gallons; and so on. Theoretically, we would never use up the entire amount of oil. 
However, the limit of the sum of the amounts of oil reclaimed is approximately 
reached after a large number of operations. Hence, we have an infinite geometric 

progression in which a = 100 and r = 4/5. Then S = jyz = 500. Thus, by 

re-refining the oil as it is used, the fleet owner has had the equivalent of 500 gallons 
of oil. 

15-12. REPEATING DECIMALS 

If a decimal contains a fixed sequence of digits which are 
repeated indefinitely, we call it a repeating decimal. Thus, 
0.135135 • • is a repeating decimal. This decimal is written 0.i35, 
the dots indicating the first and last digits of the sequence which is 
to be repeated. Also, 0.34516 = 0.34516516516 • • • . A repeating 
decimal is wholly or partly an infinite geometric series. For 



270 Progressions Sec. 15-12 

example, since 0.34516 = 0.34 + .00516 + 0.00000516 + ••• , it is 
composed of the decimal 0.34 and an infinite geometric series in 
which a = 0.00516 and r = 0.001. Hence, since \r\ < 1, 

n qakia -hqa± 000516 _ ft Q , , 0.00516 _ 34 172 _ 11494 
Ud451b " Ud4+ l -0.001 "" Ud4+ 0.999 " 100 + 33300 "~ 33300 ' 

Note that if we divide 172 by 33300, we obtain the repeating deci- 
mal 0.00516. 

Example 15-13. Express the repeating decimal 0.263 as an equivalent numerical 
fraction. 

Solution: We can write this decimal in the form 

0.2 + 0.063 + 0.00063 + • • • . 
Hence, the required number consists of the decimal 0.2 plus an infinite geometric 
series in which a = 0.063 and r = 0.01. The sum of the series, since \r\ < 1, is 

0.063 0.063 7 



8 = 



1 - 0.01 0.99 110 



1 7 9Q 

Therefore, 0.263 = ± + jjg = f± • 

EXERCISE 15-5 

In each of the problems from 1 to 12, find the sum of the convergent series based 
on the given infinite geometric progression. 

1. ^ > — 1, g ) • • • . 2. g > - > 2= > • • • . 3. 3, \/3, 1, • • • . 

4. — * -r > ~ i • • • • 5. 8, 4, 2, • • • • 6. 8, — 4, 2, • • • • 

24 a 112 q i 3 9 

10. 0.5, 0.05, 0.005, • • • . 11. 0.18, 0.0018, 0.000018, 

12. 0.3 + 0.012 + 0.00012 + 0.0000012 + • ■ • . 

13. Find the sum of the series based on the infinite geometric progression 24, 

8 
8, 5 ) - • • . Also, find the sum of the first 20 terms of this progression and com- 
o 

pute the error introduced by using S instead of $ 2 o. 

14. What would be the error if S were used instead of S n for the sum of the geometric 
progression 48, — 36, 27, • • • to 10 terms? 

15. A ball is dropped from a height of 3 feet. On each rebound it bounces back to 
three-fourths the height from which it last fell. Assuming that this bouncing 
continues indefinitely, find the distance it travels in coming to rest. How far 
has it traveled after bouncing ten times? 

16. A swinging pendulum will gradually come to rest as a result of friction. If, on 
each upswing, the pendulum swings through 98 per cent of the arc through 
which it fell, and if the initial arc for one complete swing was 20 inches, find the 
distance traveled before the pendulum comes to rest. 



Sec. 15-13 Progressions 271 

Convert each of the following repeating decimals to fractional form. 



17. 0.1. 


18. 0.15. 


19. 0.90. 


20. 0.243. 


21. 0.16. 


22. 0J42857. 


23. 2.9. 


24. U234, 


25. 0.11542. 


26. 2.i23. 







15-13. THE BINOMIAL SERIES 

We shall now consider the binomial expansion when n is any real 
number. If we let a = 1 and 6 = x in (4-13), the binomial formula 
becomes 

(15-10) (i+ g ). = i + M + 5^* + 

n(n - 1) ( n - 2) ■ ■ ■ (n - r + 1) 

+ • • • H -; x r + • • • . 

r! 

The right member of (15-10) is called a binomial series. 

We saw in Section 4-6 that, if n is a positive integer, the series 
on the right in (15-10) terminates with x n , and (15-10) is true. 
Otherwise, the terms of the series continue indefinitely, giving rise 
to an infinite series. 

The question which now arises is whether (15-10) is valid when 
n is not a positive integer ; that is, whether the series on the right 
converges, and, if so, whether its sum is equal to (1 + x) n . It is 
proved in the study of series that (15-10) is indeed valid if \x\ < 1. 
It follows that, if \x\ < 1, we may obtain the value of (1 + x) n as 
accurately as we please by taking sufficiently many terms of the 
series in (15-10). 

The expansion can readily be extended to (a + b) n when n is not 
a positive integer. In this case, the expression may be written as 
follows : 

(15-11) (a + 6)- = [a(l+J)]"=«-[l+|]'. 

Here x = - ? and the expansion of (a + b) n is valid if - < 1. 
a \a\ 



Example 15-14. Find the first five terms of the expansion of (1 + x)~ 3 if | x | < 1. 
Solution: By (15-10), 

(1 + *)-» = 1 + (■*■ 3)* + ( ~ 3) 2 f~ 4) s* 

(-3) (-4) (-5) '. (-3) (-4) (-5) (-8) 

+ 3! x + 4| x + 

= 1 - 3x + 6x 2 - 10x 3 + 15x* + • • • . 



272 Progressions Sec. 15-13 

Example 15-15. Find -^L04. 

Solution: <i/TM = (1 + 0.04) ^ Hence, n = 1/3 and x = 0.04, and the ex- 
pansion is 

(1 + 0.04)"» = 1+| (0.04) + 2! (0.04)2 + 3 ^ 3 ^ *' (0.04)3+ .... 

The approximate value of the sum of this series is 

1 + 0.013333 - 0.000177 + 0.000004 = 1.013160. 

We have here a case of an alternating series, that is, a series in which the terms 
are alternately positive and negative. It is proved in the study of series that, if in 
an alternating series each term is numerically less than the preceding term and 
lim a n = 0, the error introduced by using S n as the sum of the series is 
numerically less than the value of the first term omitted; that is, \S» - 8\ < |a»+i|. 
If in the present example we take for S the value S 3 = 1 + 0.013333 - 0.000177 
= 1.013156, the error in so doing is less than the value of the fourth term 0.000004. 

Example 15-16. Find the first four terms of the expansion of — 



V8 -x 2 

Solution: First we apply (15-11) to convert the given expression to a suitable 
form, as follows: 

(-r2\-l/3 1 / 3.2V -1/3 

^t) =10 -I) • 

Hence, by (15-10), if a; 2 < 8, 

_L_ = lfl | -1/3 / *\ , (-1/3) (-4/3) / g\» 
■ViT^i* 2\ 1 V 8 J + 1-2 V 8/ 

| (-1/3) (-4/3) (-7/3) / g\» H \ 

L * 2 ' o V o / / 



2 V ^ 24 ^ 288 ^ 20,736 ^ / 



EXERCISE 15-6 

In each of the problems from 1 to 10, find the first four terms of a binomial 
expansion of the given expression. 
1 « 1 



1 + x 1 - x 



3. VI + x. 4. VI - x. 5. (1 + x)-"*. 



6. (i - x)-i/2. 7. -4— • 8. 7— r-vi • 9. 7 ,* x, ' 10. (1 + xy>*. 

x + 2/ (a + 1/) 2 (a; + y)* v y 

Find the approximate value of each of the following numbers by means of a 
binomial expansion, using four terms of the expression. 

11. (1.02)". 12. (1.01) 13 . 13. (1.04) 8 . 14. (l.l) 10 . 15. (0.98) 8 . 

16. 49 4 . 17. (0.99)«. 18. 51*. 19. (1.03) 1 ' 2 . 20. (0.97)" 2 . 



16 



Mathematical Induction 



16-1. METHOD OF MATHEMATICAL INDUCTION 

When a certain type of formula or proposition has been verified 
in specific cases but is not known to be true in general, the method 
of mathematical induction is often found extremely valuable in 
determining its validity. 

Suppose that a statement involving a positive integer n is to be 
proved true for all values of n greater than or equal to a particular 
initial value. We begin by showing the result to be true for the first 
value of n. We then assume that k is some particular integral value 
of n for which the statement holds. With this assumption as a 
basis, we establish the validity of the statement in the next suc- 
ceeding case, namely, that in which n — k 4- 1. In other words, we 
prove that if the statement is true for any specific integral value 
of n, say n = k, then it is also true for the next larger value of n, 
namely, n = k + 1. Suppose for example, that 1 is the initial value 
of n. Then the second step establishes that if the statement is 
true for n = 1, it is also true for n = 1 + 1, or 2 ; if it is true for 
n = 2, it is also true f or n = 2 + 1, or 3 ; and so on. As a consequence 
of this, we conclude that the statement is true for all values of n 
greater than or equal to the initial value, here 1. A proof by 
mathematical induction, therefore, consists of two parts and a 
conclusion. 

Part 1. Verification that the statement is true for some initial, 
value of n, generally n = 1. (This initial value is the smallest value 
of n for which the statement is to be proved true.) 

Part 2. Proof that whenever the statement is true for some par- 
ticular value of n, say for n = k, then it is true for the next larger 
value of n, that is, for n = k + 1. 

273 



274 Mathematical Induction Sec. 16-1 

Conclusion. If both parts of the proof have been given, then the 
statement is true for all positive integral values of n greater than 
or equal to the one for which the verification was made in Part 1. 

The reasoning process involved here, which consists in taking 
an initial integer and then repeatedly taking successors, can be 
exemplified in terms of climbing a ladder. Part 1 puts us on the 
bottom rung of a ladder. Part 2 shows us how to get from any rung 
we have reached to the next higher rung. The conclusion states 
that if we know how to get on the bottom rung of the ladder, and if 
we know how to get from any rung to the next higher one, then we 
can reach all rungs, and hence can climb the ladder. 

The following examples will illustrate the method. 

Example 16-1. If n is any positive integer, prove that 
(16-1) 1+-1-+J-- + ...+ 1 - n 



l-22-3^3-4 n(n + 1) n + 1 

Solution: 

Part 1: The formula is true when n = 1, since 

1 _ 1 11 

1 • 2 "~ 1 + 1 ' ° r 2 2* 
Part 2: Let k represent any particular value of n for which (16-1) is true. Then 

(16-2) Y72 + 2T3 + 3T4 + * " * + k(k + 1) = F+T ' 

We now wish to prove that (16-1) is true also for the next larger value of n, 
namely, n = k + 1. The sum on the left in (16-1) when n = k + 1 can be obtained 

by adding its last term, which is ,, 1W , v > to both sides of (16-2). Hence, 

, v/c + l) {ic + Z) 

we have 



U-2 + 2-3 + 3-4i + *(* + !)/ ^ (& + !)(/ 



1) (* + 2) 

k +■ ' 



k + 1 (* + 1) (* + 2) 
k 1 k 2 +2^ + 1 & + 1 Jfe + 1 



But, since k + l , ( t + 1} ( t + 2 ) " (fc + l) (* + 2) " t + 2 " (* + 1) + 1 ' 
we obtain 

uo-a; i.2 i "2-3" t "3-4' t """" r "(* + l)(t+2)"(t+l)+l 

The members of (16-3) are the same as those of (16-1) when n = k -f 1. Hence, 
we have shown that (16-3) is true if (16-2) is true, in other words, (16-1) is true 
for n = k + 1 if it is true for n = k. 

Conclusion: We have shown by verification that (16-1) is true when n = 1. 
Therefore, since (16-1) is true for n = 1, it follows from Part 2 that (16-1) is true 
for every positive integer n. 



Sec. 16-2 Mathematical Induction 275 

Example 16-2. Prove that (x - y) is a factor of (r» - y 11 ) if n is any positive 
integer. 

Solution: 

Part l:\ln = 1, then x n — y n = x — ?/, which is seen to have (x — y) as a factor. 

Part 0: Let A: be a specific value of n for which (x n — y n ) has (# - y) as a factor. 
We shall now prove that if {x — y) is a factor of (a;* — ?/*), it is also a factor of 
( x *+i - y*+i). We have 

z*+i _ y* +1 = z* +l - xy* + zy* - y k+1 = x(x* - y k ) + */*(# ~ !/)• 

By assumption, (x — ?/) is a factor of (x k — y k ). Also, by inspection, (x — y) is a 
factor of t/*(x - y). Hence, (x - y) is a factor of the left member (x* +1 - y k+1 ). 
Therefore, if the conclusion is true for n = A*, it is also true for n = A; + 1. 

Conclusion: By Part 1 of the proof, (x - y) is a factor of (x n - ?/ n ) when n = 1. 
Therefore, by virtue of Part 2, the desired conclusion is true for any positive 
integer n. 

16-2. PROOF OF THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENTS 

We shall now prove that the binomial formula, 
(16-4) (a + &)» =a n + na n ~ l b + n{jl ~ ^ a n ~ 2 b 2 

n(n-l)---(n-r + 2) ^^ 

(r - 1)! 

, n(n — 1) • • • (n — r + 1) , . 

-| — >. ^ p £ a n r b r + • • • + o w > 

r! 

is true for every positive integral value of n. 

Proof. Part 1. When n = 1, each side of (16-4) becomes a + &. 
Hence, (16-4) is true for n = 1. 

Par£ #. Let /c be any specific value of n for which (16-4) is true. 
Then we have 

(a + b) k = a* + ka k ~ l b + k<<k ~ ^ a k ~ 2 b 2 

r! 
Multiplying each member of this equation by (a + b), we obtain 

( a + 6 )*+i = a *+i + fca *& + . . . + * ( * = 1} - \ (fc = r + 1} a*-'* 1 **' 

r\ 

+ • • • + ab k + a*6 + • • • 

+ W-l)...(fe-r + 2) ^ r+1&r + _ + fct+1 

(r - 1) ! 



276 Mathematical Induction See. 16-2 

Hence, by combining terms, we get 

(o + b) k+1 = a* +1 + (Jfc + l)a*6 + • • • 

(ib + l) t ...(fc-r + 2) ^^ ftt+1 

r! 

In this result the sum of the coefficients of a k - r+1 b r is obtained as 

follows : 

k( k - 1) - --(ft -^r + 2) (fe - r + 1) fc(Jfc - 1) . . . (Jfe - r + 2) 



r! 



-P^±! +.].[: 



(r-1)! 

fc(Jb - 1) ; ; ; (fc - f + 2)' 



r J l (r-1)! J 

_ (fc + 1) (fc) (fc - 1) • » ■ (Jb - r + 2) 
r! 

We note that the value of (a + b) k+1 , obtained as the product of 
(a + &)* and (a + 6), is exactly the same as the expansion which 
would be obtained from (16-4) with n = fc + l. Hence, we have 
shown that if the binomial formula holds for n = k, it must hold for 
n = fc + 1. 

Conclusion. The binomial formula was seen to be true for n = 1. 
Therefore, by virtue of Part 2, we may conclude that it is true for 
every positive integer n. 



EXERCISE 16-1 

Prove by mathematical induction that each of the statements in Problems 1 to 
20 is true for all positive integral values of n. 

, n(n + 1) 



1. 1 + 2 + 3 + • 

2. 1 + 3 + 5 + • 

3. 2 + 4 + 6 + • 

4. 1 + 3 + 6 + • 

5. 3 + 6 + 9 + • 

6. 4 + 8 + 12 + 

7. 1 + 4 + 7 + • 

8 i+i + i + . 
8 * 2 + 4 + § + 

9. 1 • 2 + 2 • 3 + 3 
10. 1* + 2* + 3* + 



+ n =■ 

4- (2n - 1) = n 2 . 
+ 2n = n(n + 1). 

n(h + 1) _ n(n + 1) (n + 2) 
2 ~~ 6 



+ 



+ 8n =§2fiLtl). 

• +4n =2n(n + 1). 

+ (3n -2)=^L=J). 

n . _ «(n + 1) (an + 1) , 



Sec. 16-2 Mathematical Induction 277 

11.2»+4»+6»+--- + (2n)'= 2w(w + 1 > (2w + 1) . 

o 

12 . P+23+33 + . .. +n . = [2fi^tiiJ. 

13. I 3 + 3 3 + 5 3 + • • • + (2n - l) 3 = n 2 (2r* 2 - 1). 

14. 2 3 + 4 3 + 6 3 + • • • + (2n) 3 = 2n 2 (n + I) 2 . 

15. 2 + 2 2 + 2 3 + • • • +2« = 2(2» - 1). 

16. 3 + 3 2 + 3 3 + • • • + 3» = | (3» - 1). 

17. 4 + 4 2 + 4 3 + • • • + 4» = | (4» -,1). 

18. I 2 + 3 2 + 5* + 7 2 + • • • + (2n - | 2 = n(2n - 1) gn + 1) . 

19 ±, 1 , 1 , . 1 _Vn 



1-3 ' 3 • 5 ^ 5 • 7 ^ ^ (2n - 1) (2n + 1) ~ 2n + 1 

20 1 i 1 i 1 i , * = n(n+3) . 

1 • 2 • 3 ^ 2 • 3 • 4 **" 3 • 4 • 5 ^ "*" n(n + 1) (n + 2) 4(n + 1) (n + 2) 

21. Prove that # 2n - y 2n is divisible by a: -f- y for every positive integer n. 

22. Prove that a; 2 * 1 - 1 + y 2n ~ l is divisible by x + 2/, for every positive integer n. 

23. By using mathematical induction, prove the formula for the sum of an arith- 
metic progression. 

24. By using mathematical induction, prove the formula for the sum of a geometric 
progression. 



17 



Permutations, Combinations, 

and Probability 



17-1. FUNDAMENTAL PRINCIPLE 

We begin the study of permutations and combinations by con- 
sidering the following principle, which is fundamental for the 
entire subject. 

Fundamental Principle. If one thing can be done in a ways, and 
if, for each such way, a second thing can be done in b ways, then 
the two together can be done in a • b ways. 

To understand why the principle is true, note that for each of the 
a ways of doing the first thing, there are b ways of doing the 
second ; hence, both the first and the second things taken together 
can be done in a • b ways. 

The following examples will illustrate the reasoning upon which 
the principle is based, as well as an obvious extension of the prin- 
ciple to the case when more than two things are to be done. 

Example 17-1. In how many ways can two officers, a chairman and a secretary, 
be selected from a committee of five men? 

Solution: By the fundamental principle, the problem is equivalent to determining 
the number of ways in which the two things can be done together. The first 
position can be filled in 5 ways; that is, there are a = 5 ways of selecting a chairman. 
For each of these possible selections, there are b = 4 ways of filling the position of 
secretary from the remaining men. Hence, the number of ways of selecting a 
chairman and secretary is a • b = 5 • 4 = 20. 

Example 17-2. How many three-digit numbers can be formed from the ten 
digits 0, 1, 2, • • • , 9, if a) repetitions of digits are not permitted; 6) repetitions 
are permitted? 

Solution: a) Here we have three things to do or places to fill. The hundreds 
place can be filled in 9 ways, since must be excluded from this place. The tens 

278 



Sec. 17-2 Permutations, Combinations, and Probability 279 

place can then be filled in 9 ways from any of the remaining 9 digits. Finally, the 
units place can be filled from any of the remaining 8 digits. There are, therefore, 
9.9.8 = 648 three-digit numbers in which no two digits are alike. 

b) If repetitions are permitted, there are 9 ways to fill the hundreds place and 
10 ways to fill each of the tens and units places. Hence, there are 9 • 10 • 10 = 900 
three-digit numbers. 

EXERCISE 17-1 

1. Nine persons apply for each of two vacant apartments. In how many possible 
ways can both apartments be rented? 

2. A large room has eight doors. In how many ways can a person enter the room 
by one door and leave by a different door? 

3. If three dice are thrown, in how many ways can they fall? 

4. There are eight men and six women in a club. In how many ways can two 
officers be selected so that one is a man and one is a woman? 

5. How many possible four-digit numbers are there in a telephone exchange which 
uses only four-digit numbers and excludes 0000? 

6. How many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6, 7, 8, 9? 
How many of these are larger than 700,000? 

7. How many numbers of six different digits can be formed from the digits 2, 3, 4, 
5, 6, 7, 8, 9? How many of these are larger than 700,000? 

8. A woman has seven guests at a party. If she chooses her seat first, in how many 
ways can she seat her guests? 

17-2. PERMUTATIONS 

An ordered arrangement of all or any part of a set of things is 
called a permutation. Specifically, suppose we have n distinct things 
and wish to select r of these to be arranged in a definite order. 
Each such ordered arrangement is called a permutation of n things 
r at a time. The number of all such permutations is denoted by 
n P r . Thus, 5 P 2 is read "the number of permutations of 5 different 
things 2 at a time." 

In Example 17-1, the possible number of ways of selecting a 
chairman and secretary from a committee of five men was found 
by means of the fundamental principle to be 5 • 4 = 20. This is pre- 
cisely equal to 5 P 2 , since it is the number of ways in which two men 
can be chosen from among the given five men and arranged in the 
two offices. The number of permutations of n things rata time is 
given by the formula 

(17-1) n P r = n(n - 1) (n - 2) • • • (n - r + 1). 

The truth of (17-1) is readily shown as follows. The first of the 
r places can be filled in n ways. Then the second can be filled in 
(n — 1) ways, the third in (n — 2) ways, and so on. In general, the 



280 Permutations, Combinations, and Probability Sec. 17-2 

number of ways of filling each place is n minus the number of places 
already filled. Therefore, when the rth object is chosen, (r — 1) 
places have already been filled, and the rth place can then be filled 
inn- (r — 1) , or n - r + 1, ways. 

In particular, if r = n, the last factor becomes n — n + 1 = 1. We 
then have 

(17-2) n P n = n(n - 1) (n - 2) • • • 1 = n!. 

This formula gives the number of permutations of n different 
things taken all at a time. 

If both the numerator and the (understood unit) denominator 
of (17-1) are multiplied by (n — r) !, we obtain the following alter- 
nate formula : 

(17-3) P - n ( n ~ *) ( n ~ 2 ) ' ' * ( n ~ r + *) ' ( n ~ r ) ! = n! 
^ ' n r (n — r)\ (n — r)\ 

Here it is agreed that by definition ! = 1. Hence, (17-3) holds for 
r = n. 

For example, by (17-2), 
8 P 6 = n(n - 1) (n - 2) • • • (n - r + 1) = 8 • 7 • 6 • 5 • 4 = 6720. 
Using (17-3), we have also 

j, n! 8! 8-7-6-5-4-3-2-1 A7on 

•ft = (S-zt^i = si = F2^ = 672a 

17-3. PERMUTATIONS OF n THINGS NOT ALL DIFFERENT 

Suppose it is required to find the number of indistinguishable 
distinct permutations of n things all at a time, if w x things are 
regarded as indistinguishable, n^ other things are regarded as indis- 
tinguishable, and so on. Let us consider, for example, the number 
of permutations of the letters a, a, a, b taken four at a time. For 
convenience, indistinguishable objects are given the same notation. 

Denote by P the desired number of distinct permutations. Evi- 
dently, P is less than 4P4, which is the number of permutations that 
could be effected if all the letters were distinguishable. For in any 
one of the P permutations, say (a b a a) , any rearrangement of the 
a's among themselves would not change the permutation. If, how- 
ever, we assigned subscripts to a in this permutation, as in 
(a 2 6 a® a 3 ), we could permute these three distinct letters among 
themselves in 3 ! ways. This can be done for each of the P permu- 
tations of the letters a, a, a, b. We would then obtain P • 3 ! per- 
mutations of the four distinct letters a u ch, a 3 , b taken four at a 
time. There are therefore 4 ! permutations altogether. Hence, 

4f 4! 
P. 8! =4!, or P = |j = %l 



3! 3!1! 



Sec. 17-4 Permutations, Combinations, and Probability 281 

In general, the number of distinct permutations of n things taken 
all at a time, if n x things are alike, n 2 other things are alike, n 3 other 
things are also alike, and so on, equals 

n! 



(17-4) P = 



17-4. COMBINATIONS 



n\\ ri2\ n$\ 



A combination is a set of all or any part of a collection of objects 
without regard to the order of the objects in the set. We use the 
symbol n C r to represent the total number of all combinations of n 
different things taken r at a time. 

The different sets of the four letters a, b, c, d taken three at a 
time, without reference to the order in which the letters are 
arranged, are (abc), (abd), (acd), (bed). From each of these four 
combinations, we can form 3 !, or 6, different permutations of the 
four letters taken three at a time. For example, from the combina- 
tion (abc) we can form the distinct permutations (abc), (acb), 
(bac), (bca), (cab), (cba). Hence, each of the four combinations 
contributes 3 P 3 = 3 ! permutations to the total number of permu- 
tations. Thus, there are 4 C 3 combinations of the four letters if we 
disregard order, and there are 3! ordered arrangements or per- 
mutations of each combination. Hence, we have 4 P 3 = 3 ! • 4 C 3 , or 

4 C 3 = ~y = ' ' ' = 4 combinations. This is the same as the 

o! 1 • 2 • o 

number we obtained at the beginning of the paragraph. 

In general, if we divide the total number of permutations of n 

things rata time, or n P rp by the number of permutations, or r!, 

contributed by each combination, we obtain the total number of 

combinations. Symbolically, we have 

or 

(17-5) n C r = „, 



P 



If we note that r ! = r P r , then we can write the following inter- 
esting relationship : 

o — n r . 

nW — p 

Replacing n P r by its equivalent expression from (17-1) or (17-3) 
we have 

(17-6) nC r 7{ r!(n-r-r)! 



282 Permutations, Combinations, and Probability Sec. 17-4 

— r) in 



If r is replaced by (ft — r) in (17-6) , we obtain 

n! 



(n — r)! r! 
Hence, 

Note. Having agreed that ! = 1, we can supply (17-1) or (17-3) 
to permutations or combinations of n things zero at a time. Con- 
sidering (17-5), we have 

n! 1 

(17 ~ 8) nCo = 0!(ft-0)! = 0! = L 

This result agrees with the intuitive conclusion that there is only 
one such "empty" combination. Similarly, n P = 1. 

17-5. BINOMIAL COEFFICIENTS 

By referring to the development of the binomial formula in Sec- 
tion 4-6, we note that the expansion of (a + b) n involves the prod- 
uct (a + b) (a + b) (a + b) • • • taken without regard to order. This 
fact suggests the use of combinations in the coefficients of the 
expansion. 

We said in Section 4-7 that the coefficient of the term involving 

a n ~ r b r is — - - - - . ~~ r • This is precisely the formula for 

r! 

the combination of n things rata time, or n C r , obtained in Section 
17-4. The binomial formula may therefore be written as follows : 

(17-9) (a+b)» = n Coa«+ n C 1 a n - l b+ n C 2 a n - 2 b 2 +. • .+ n C r a-^+. . -+ n CJ>\ 

For example, the expansion of (a + b) 4 takes the following form: 

4 C a 4 + 4 Cia 3 5 + 4 C 2 a 2 6 2 + *Csab 3 + 4 C 4 6 4 . 

This reduces to 

a 4 + 4a 3 b + da 2 b 2 + 4ab 3 + b 4 . 

If we let a = b = 1 in the expansion for (a + b) n in (17-9), we 
obtain 

(1 + 1)" = n C + nCl + n C 2 + ' • • + nC n . 

Since n C = 1, n Ci + n C 2 + • • • + n C n = 2* - 1. 

Thus, the total number o;f combinations of n things taken succes- 
sively 1, 2, • - - , n at a time is 2* — 1. 

EXERCISE 17-2 

1. Evaluate 5P2, 7P3, 12^6, 21P4, 100F3. 

2. Evaluate 10C4, 11C3, iooCsj 21C5, iooCos* 



Sec. 17-6 Permutations, Combinations, and Probability 283 

3. Form all possible distinct permutations of the letters of the word theory, 
a. How many are there? b. How many begin and end with a vowel? c. How 
many begin or end with a vowel? 

4. How many distinct permutations are there of the five letters a, b, c, d, e taken 
three at a time? Write them out. 

5. In how many different ways can a dime, a quarter, and a half-dollar be dis- 
tributed among five boys? 

6. How many different sums can be formed with a penny, a nickel, a dime, and 
a quarter? 

7. Expand (a -f 6) 8 , using combination symbols. 

8. From the six digits 1, 2, 3, 4, 5, 6, form all permutations taken five at a time. 
a. How many are formed? b. How many begin with 4? c. In how many does 
the digit 3 not appear? 

9. How many distinct permutations can be made from the letters of the word 
probability taken all at a time? 

10. How many different combinations are there of 4 identical nickels ; 5 identical 
dimes, and 6 identical quarters? 

11. How many different straight lines are determined by twelve points, no three 
of which are in a straight line? 

12. In how many different ways can a student select seven questions out of ten 
on a test? 

13. How many different weights can be formed with six objects weighing 1, 2, 4, 8, 
16, and 32 pounds, respectively? 

14. In how many different ways can signals be made with seven different flags, 
where a signal is a set of one or more flags arranged in a specific order? 

15. In how many different ways can the 52 cards of a bridge deck be dealt among 
four players? 

17-6. MATHEMATICAL PROBABILITY 

If, in a given trial, an event can happen in h different ways and 
can fail to happen in / ways, and if all the h + f ways are equally 
likely, then the probability p that the event will happen in this 
trial is 

(17-10) p= 1 ± ? . 

The probability q that the event will fail to happen is 

(17-11) , = r ^_. 

Note that O^p^l, ^q^l f and p + q = 1. Two illustrations of 
probability follow. 

If a bag contains 3 green marbles and 4 yellow marbles, all 
exactly alike except for color, the probability of drawing a single 
green marble is 

3 3 

V 3 + 4 7 # 



284 Permutations, Combinations, and Probability Sec. 17-6 



A die can fall in six ways. The probability of getting 5 or more 

2 1 
ith one throw of a die is - > or - > since th 

o o 

either a 5 or a 6, for the event to happen. 



2 1 

with one throw of a die is - > or - > since there are two possible ways, 

o o 



17-7. MOST PROBABLE NUMBER AND MATHEMATICAL EXPECTATION 

Let p be the probability of occurrence of some event. Further- 
more, suppose that n trials of the event are made, of which h are 

successful. Then - is called the relative frequency of success for 
n 

the trials which occurred. It is not to be expected that - = p. It is 

shown in more advanced treatments of probability, however, that if 
n is large, it is very likely that the relative frequency is approxi- 
mately equal to p. Also, the larger we take n, the more likely it is 

that - approximates p closely. Moreover, it can be shown that the 
n 

most probable or expected number of occurrences of the event for n 
trials is np. 

For example, when a coin is tossed, the probability of getting a 
head is 1/2. In 1,000 trials the expected number of heads is there- 
fore 500. This does not mean, however, that if the first 100 trials 
result in 75 heads and 25 tails, we should expect 25 heads and 75 
tails in the next 100 trials. Actually, since one toss of the coin does 
not affect the next one, we should expect about 50 heads and 50 tails 
in the next 100 trials. Moreover, we may expect about 450 heads 
and 450 tails in the next 900 trials. 

If p is the probability of winning a certain amount of money in 
case a certain event occurs and m is the amount of money to be won, 
the mathematical expectation is defined to be pm. For instance, if a 
person can win $12 provided he throws an ace with a die, his expec- 
tation is - ($12) = $2. Hence, $2 is the fair amount he should be 
willing to pay to make the trial. 



17-8. STATISTICAL, OR EMPIRICAL, PROBABILITY 

It is frequently impossible to have sufficient knowledge before- 
hand of all the conditions that might cause an event to happen or 
fail to happen. In such a case, however, it may be possible to deter- 
mine the relative frequency of the occurrence of the event from a 
large number of trials. Thus, if an event has been observed to 
happen h times in n trials, and n is a large number, then until addi- 



Sec. 17—9 Permutations, Combinations, and Probability Hl$ 

tional knowledge is available, we define the statistical probability, 
or empirical probability, to be 

(17-11) p=l> 

where - is the relative frequency. 
n 

Example 17-3. A molding machine turns out 12 parts per minute. Inspection 
experience has shown that there are 20 defective parts per hour. What is the 
probability that a single part, picked at random, will be defective? In a run of 
10,000 parts, how many defectives should be expected? 

Solution: The parts are produced at the rate of 720 per hour, and 20 of them are 
defective. Hence, the probability that a single part selected at random will be 

defective is =xr = ™ • In a run of 10,000 parts, we should expect ^ (10,000), or 

approximately 278, defective parts. 

17-9. MUTUALLY EXCLUSIVE EVENTS 

Two or more events are mutually exclusive if not more than one 
of them can happen in a given trial. The following theorem may 
be stated. 

Theorem. The probability that some one of a set of mutually 
exclusive events will happen in a given trial is the sum of the indi- 
vidual-event probabilities. 

Proof. Consider, for simplicity, a set of two mutually exclusive 
events. Suppose that the first can happen in h x ways and the second 
can happen in h 2 ways, and let n be the total number of ways in 

which the two events can happen or fail to happen. Then p\ = — 

lb 

and p2 = — are the corresponding probabilities of the two events. 
Tt 

Since the n events are mutually exclusive, the h x ways are differ- 
ent from the h 2 ways, and the number of ways that either the first 
or the second event can happen is therefore hi + th. Hence, the 
probability p that either the one event or the other will happen is 

,+ *, 10N hi + hi h\ h 2 

(17-12) p= ___ = _ + _ = pi -r p2 . 

For example, suppose that a bag contains 2 green marbles, 3 yel- 
low marbles and 5 brown marbles. If a marble is drawn at random, 

2 
the probability that it is green is ^ > and the probability that it is 

3 
yellow is — • Hence if a marble is drawn, the probability that it is 

.., „ .2,3 51 

either green or yellow is — + ^ > or ^ > or ^ • 



286 Permutations, Combinations, and Probability Sec. 17-10 

17-10. DEPENDENT AND INDEPENDENT EVENTS 

In case two or more events are not mutually exclusive, they are 
dependent if the occurrence of any one affects the occurrence of the 
others, and they are independent if the occurrence of one does not 
affect the occurrence of the others. 

For example, if a card is drawn from a deck of 52 cards and the 
card is not replaced before a second is drawn, then the second 
drawing is dependent on the outcome of the first. If, however, the 
first card is replaced, then the second drawing is independent of 
the first. In the latter case the two drawings are equivalent to 
simultaneous drawings from two decks. 

We shall now state and prove the following theorem relating to 
dependent and independent events. 

Theorem. The probability that two dependent or independent 
events will occur (successively if dependent; successively or simul- 
taneously if independent) is the product of their individual 
probabilities. 

Proof. Suppose that the first event can happen in h x out of a 
total of n x different ways, and that the second event can happen 
in hz out of n*> different ways. Then it follows, by the fundamental 
principle in Section 17-1, that the two events can happen together 
in hxh 2 ways out of a total of n x n 2 different ways. Therefore, the 
probability that both events will happen is 

(17-13) p = = — • — = p\P2> 

U\U2 U\ n 2 

Example 17-4. Two cards are drawn from a deck containing 52 cards. Find the 
probability that both cards are aces when the first card is not replaced before the 
second is drawn. 

Solution: We shall begin with a listing of the following useful probabilities: 

4 

1) The probability of drawing an ace from a deck of 52 cards is ~ • 

uZ 

2) If the first card is an ace and it is not replaced, the probability of drawing 

another ace is -=^ • 
ol 

3) If the first card is not an ace and is not replaced, the probability of the 

second being an ace is — • 
ol 

4 

4) If the first card is replaced, the probability of the second being an ace is ^ • 

This drawing is entirely independent of the first drawing. 
Consider, now, the given problem. The probability that the first card is an ace is 
4 1 
P l = ^9 = 13 " ^ ^ e ^ rs ^ carc * drawn * s an ace > ^ nen the probability that the 



Sec. 1 7-1 1 Permutations, Combinations, and Probability 287 

3 1 
second card drawn is an ace is P2 = ~y = j~ • Hence, the probability that both 

•ii u • 111 

will be aces is pip 2 = 13 ' 17 = 221 * 

17-11. REPEATED TRIALS 

Theorem. If p is the probability that an event will happen in any 
trial, and q = 1 — p is the probability that it will fail, then the prob- 
ability that it will happen exactly r times out of n trials is 

(17-14) nCrP'q*-* = f{ fr]_ f)[ PT^ 

Proof. The happening of the event in exactly r trials and its fail- 
ure in the remaining n - r times are independent events. Hence, 
the probability, by the theorem in Section 17-10, is p r q n ~ r . But these 
r trials can be chosen from the n trials in n C r ways. Since these 
ways are mutually exclusive, the total probability is n C r p r q n ~ r * Note 
that this expression is the (r + l)th term of the binomial formula 
for (q + p) n , since 

(q + p)» = q" + nCiq^p + n C 2 q n ' 2 p 2 + h nC r q n - r p r + • • • + p n . 

The successive terms of this expansion give the probabilities that 
the event will happen exactly 0, 1, 2, • • , r, • • • n times in n trials. 

An event will happen at least r times in a given number of n 
trials if it happens n, n — 1, • • • , r + 1, or r times. Since these events 
are mutually exclusive, the probability that an event will happen at 
le&st r times is given by the sum 

P« + n C n _l qp"" 1 + --'+nC r q n ~ r p\ 

Example 17-5. What is the probability of tossing an ace exactly three times in 
four trials with one die? 

Solution: Since the probability of tossing an ace in one trial is ^ and the prob- 
5 
ability of failure is -r 1 we may substitute in the term n Cr5 rn " r P r of the binomial 

formula. The result is 

4C7s (DXI)^ 4 (I) (lie) = sli* 

Example 17-6. What is the probability of tossing an ace at least twice in four 
trials with one die? 

Solution: The event will happen at least twice if it happens 4, 3, or 2 times. 
Hence, the probability is given by the following sum : 

•Mi)'W!)a)'wf)'a)'=ir 



288 Permutations, Combinations, and Probability Sec. 17-1 1 

EXERCISE 17-3 

1. A certain event can happen in four ways and can fail to happen in six ways. 
What is the probability that it will happen? If $60 can be won on the event, 
what is the mathematical expectation? 

2. A box contains 5 white balls, 4 red balls, and 13 black balls, a. If one ball is 
drawn out, what is the probability that it is red? b. What is the probability 
that it is white or red? 

3. a. If one die is thrown, what is the probability that a "1" or a "2" will turn up? 
b. What is the probability that a "3" or larger number will turn up? 

4. When a coin was tossed 100 times, 80 heads and 20 tails turned up. If the 
tossing were continued until 200 tosses had been made, what would be the 
most probable number of tails in the second 100 tosses? 

5. A bag contains five $1 bills, ten $5 bills, and twenty $10 bills. If one bill is 
drawn, what is the mathematical expectation? 

6. What is the probability of throwing a "7" or an "11" on one throw of two dice? 

7. An automobile owner carries $1,000 theft insurance on his car. If, during the 
past year, 237 out of 97,864 automobiles registered in his area were stolen, 
what is the mathematical value of the policy? 

8. In a city of 77,000 families, a careful sample of 800 families showed that 120 
of the sample families owned their own homes, a. What is the probability 
that a family selected at random in the city owned its home? b. What is the 
expected number of families in the city who own their own homes? 

9. In a certain city 28,600 persons voted for one candidate for an office, and 
23,100 voted for his opponent. What is the probability that a voter chosen at 
random voted for the winner? 

10. A bag contains 5 red balls and 9 black balls. If two balls are drawn in suc- 
cession, and the first is not replaced, find the probability that the first is red 
and the second is black. 

11. In a baseball tournament the probability that team A will win is = > and the 

1 
probability that team B will win is ^ • Find the probability that one of these 

two teams will win. 

2 

12. The probability that team A will reach the finals of a tournament is » > 

1 7 

and the probability that it will win the finals is - • Find the probability that 

team A will win the tournament. 

13. Find the probability of throwing three successive fours on a pair of dice. 

14. The probability of A winning a game when he plays it is j • He is scheduled 
to play four times, a. Find the probability that he will win exactly three times. 
b. Find the probability that he will win at least three times. 

15. Three dice are tossed, a. Find the probability that exactly two threes will 
turn up. b. What is the probability that at most two threes will turn up? 



18 



Solution of the 
General Triangle 



18-1. CLASSES OF PROBLEMS 

There are certain relationships among the lengths of the sides 
and the trigonometric functions of the angles of every triangle. If 
one side and any two other parts of a triangle are given, the 
remaining parts can be determined; that is, the triangle can be 
solved. The three given parts may comprise any one of the follow- 
ing four combinations : 

Case I. One side and two angles. 

Case II. Two sides and the angle opposite one of them. 

Case IIL Two sides and the included angle. 

Case IV. Three sides. 
In this chapter, we shall discuss methods for treating these four 
cases. For convenience, we shall let ABC denote any triangle 
whose angles are A, B, and C; and we shall let a, b, and c represent 
the lengths of the corresponding opposite sides. 

18-2. THE LAW OF SINES 

Law of Sines. Let ABC be any triangle lettered in the conven- 
tional manner. Then the following relationship between the sides 
and the sines of the angles may be written : 

(18-1) « = * = «. 

sin A sin B sin C 

This relationship is commonly called the law of sines. 

c 



B A 



Fig. 18-1. 
289 





290 Solution of the General Triangle Sec. 1 8-2 

Proof. We first note that all angles may be acute, as in Fig. 
18-1 (a), or one angle, say B, may be obtuse, as in Fig. 18-1(6). 
(The case where B = 90° entails no difficulties and will therefore 
be omitted.) In each diagram, let h denote the altitude from 

the vertex C to the side AB. Then, in either case, sin A = r and 

h 
sin B = - • Dividing the first equation by the second, we have 
a 

sin A a a b 

or 



sin B b sin A sin B 

In a similar way, by drawing the altitude from the vertex A to 
the side BC, we get 

b c 

sin B ~~ sin C 

The equations thus obtained may be combined to give the law of 
sines 

sin A sin B sin C 

Note. The law of sines is well adapted to the use of logarithms 
because it involves only multiplications and divisions. 

Since any pair of ratios in the law of sines involves two angles 
and the sides opposite, it may be used in the solution of problems in 
Cases I and II. 

As noted above, the law of sines also applies in the special case 
where ABC is a right triangle. In this case, one of the angles is 
90°, and the sine of that angle is 1. 

18-3. SOLUTION OF CASE I BY THE LAW OF SINES: GIVEN ONE SIDE AND 
TWO ANGLES 

When one side and any two angles of a triangle are known, the 
third angle can be found from the relation A + B + C = 180°, and 
each of the required sides is uniquely determined. These sides may 
be found by the law of sines. 

Example 18-1. In a triangle ABC, A = 38°14', B = 67°20', c = 329. Solve 
the triangle. 

Solution: The values of the given parts are indicated in Fig. 18-2. In this case, 
c C = 180° - (38°14' + 67°20') = 74°26'. 

To find a, we use the relationship 




a 



a 329 



sin 38°14 / sin 74°26' 

_____ Therefore, 

c-329 " 329 sin 38°14 / 

Fig. 18-2. a "" sin 74 26' ' 



Sec. 18-4 



Solution of the General Triangle 



291 



and 



log a = log 329 + log sin 38°14' - log sin 74°26\ 
The indicated operations may be performed as follows: 



log 329 = 
log sin 38°14' = 

log sin 74°26 / = 

log a = 



2.5172 
9.7916 - 10 
12.3088 - 10 
9.9838 - 10 



2.3250 
a =211. 
To determine b, we use the relationship 

329 



Hence, 



and 



sin 67°20' ~ sin 74°26' 
329 sin 67°20' 



6 =: 



sin 74°26' 



log b = log 329 + log sin 67°20' - log sin 74°26'. 
The work follows: 

log 329 = 2.5172 
log sin 67°20' = 9.9651 - 10 



log sin 74°26' = 
log b = 



12.4823 - 10 

9.9838 - 10 



2.4985 
b = 315. 
As shown in Fig. 18-1 (a), 

c = b cos A + a cos B. 
This relationship may be used as a check. Thus, 

c = 315 cos 38°14' + 212 cos 67°20' 
= (315) (0.7855) + (212) (0.3854) = 329.1, 

which agrees satisfactorily with the given value of c. 

18-4. SOLUTION OF CASE II BY THE LAW OF SINES GIVEN TWO SIDES AND 
THE ANGLE OPPOSITE ONE OF THEM 

Case II is called the ambiguous case, because the data may be 
such that two, one, or no triangles are determined. The number of 
solutions when a, b, and A are given is indicated by the accompany- 
ing table. 

Table of Possible Solutions 



(18-2) 
(18-3) 
(18-4) 
(18-5) 


A acute 


a = b sin A 
a <b sin A 
b sin A < a <b 
a^b 


One right triangle 
No triangle 
Two triangles 
One triangle 


(18-6) 
(18-7) 


A obtuse 


a^b 
a>b 


No triangle 
One triangle 



292 



Solution of the General Triangle 



Sec. 18-4 



We shall use Fig. 18-3 to illustrate in turn the different possi- 
bilities considered in the table. If the angle A and the sides a and b 
are given, we first construct the angle A with the initial ray AX 
and the terminal ray AR. Next, we lay off the distance AC = 6 
along the terminal side. Then, with C as the center and the length 
of the side a as the radius, we describe an arc. We mark the point 
or points in which this arc intersects the initial ray AX of the 
angle A. 




Fig. 18-3. 



Figure 18-3 (a) corresponds to (18-2) in the table. Since 
BC = a = b sin A, this segment is the altitude of the triangle drawn 
from the vertex C. Hence, the arc with radius a is tangent to the 
initial side at B, and the triangle is a right triangle. 

Figure 18-3(6) corresponds to (18-3) in the table. Since 
a < b sin A, the side a is too short to intersect AX, and there is no 
triangle. 

In Fig. 18-3 (c), a < b and b sin A < a, as stated in (18-4) in the 
table, and the arc will intersect AX in two points marked B and B'. 
Therefore, two solutions exist. The angle 2?' in the triangle AB'C is 
the supplement of the angle B in the triangle ABC. 

Figure 18-3 (d) represents the case in which the side a is longer 
than the side 6, as stated in (18-5) in the table. Hence, there is 
only one point B in which the arc with radius a intersects the initial 
ray AX of the angle A. There is only one solution. 



Sec. 18-4 



Solution of the General Triangle 



293 



In Fig. 18-3 (e), the angle A is obtuse. Since a < b, as stated in 
(18-6), the radius a is too short to intersect the initial ray AX, 
and no triangle exists. 

Finally, in Fig. 18-3 (/) , A is obtuse and a > b, as stated in 
(18-7). Here the arc can intersect the initial ray AX in only one 
point. There is, in this case, only one triangle. 

The following examples will illustrate some of the possibilities. 

Example 18-2. In a triangle ABC, A = 36°15', a = 9.8, b = 12.4. Solve the 
triangle. 

Solution: Draw Fig. 18-4 approximately to scale, showing the given parts. After 
angle A and side b have been drawn, an arc is described with C as center and a as 
radius. The arc intersects the side AX in two points, B\ and B 2 , and we apparently 
have two possible triangles, AB X C and AB 2 C. 




Fig. 18-4. 



To find sin B, we use 



Then 



sin B sin A 



sin B = 



b a 

12.4 sin 36°15' 



9.8 



and 

log sin B = log 12.4 + log sin 36°15' - log 9.8 
The logarithmic work follows: 

log 12.4 = 1.0934 
log sin 36°15 ; = 9.7718 - 10 
10.8652 - 10 
log 9.8 = 0.9912 
log sin B = 9.8740 - 10 
B = 48°26'. 
There are two solutions, since 6 sin A < a < b. The left inequality follows from 
the fact that log sin B = log ^*A < o, and so ^4 < 1. If we let B x = 48°26', 



294 



Sblution of the General Triangle 



Sec. 18-4 



then B 2 = 180° - 48°26' = 131°34' leads to another solution. Thus, 

B x = 48°26', Ci = 180° - (36°15' + 48°26') = 95°19', 
and 

B 2 = m^', C 2 = 180° - (36°15' + 131°34') = 12°11'. 
To find Ci, we use the law of sines again. Thus, 

9.8 sin 95°19' 



Ci = 



Similarly, we have 



c 2 = ■ 



sin 36°15' 
9.8 sin 12°11' 



16.5. 



= 3.5. 



sin 36°15' 

As a partial check, the equation Ci = b cos A + a cos Bi may be used. Thus, 
12.4 cos 36°15' + 9.8 cos 48°26' = 12.4 (0.8064) + 9.8 (0.6635) = 16.5. 
This result is the same as the value previously calculated. The same method may 
be applied to check c 2 . 

Example 18-3. Given A = 56°30 , a = 13.0, b = 10.7, solve the triangle ABC. 

Solution: Here we clearly have only one solution, since a > b. This can be seen 
geometrically if we draw Fig. 18-5 approximately to scale and show the given parts. 
Since a is greater than b, an arc with C as center intersects AX on opposite sides of 
A. Obviously there is only one triangle, ABiC, containing the angle A. 

To find J5i, we use the relationship 
sin Bt sin 56°30' 



a - 13.0 




10.7 



13 



whence 
x log sin Bi = log 10.7 

This gives 



Fig. 18-5. 
Then C = 180° - (56°30' + 43°20') = 80°10'. 
To find c, we use the law of sines and obtain 

c 13 



+ log sin 56°30' - log 13. 
B l = 43 o 20'. 



sin 80°10' sin 56°30' 



This gives c = 15.4. 



Example 18-4. Given A = 67°40', a = 16.0, b = 17.3, solve the triangle. 

C Solution: The given parts are shown in Fig. 18-6. By 

the law of sines, we have 

sinB _ sin 67°40 / 
17.3 " 16 
Therefore, 
log sin B = log 17.3 -f log sin 67°40' - log 16 

= (1.2380) + (9.9661 - 10) - (1.2041) = 0. 
Hence, B = 90°. 

The solution may be completed by applying the theory 
of right triangles. Only one solution exists. 




Sec. 18-4 Solution of the General Triangle 295 

Example 18-5. Given A = 47°23', a = 230, b = 720, solve the triangle. 

Solution: From Fig. 18-7, it appears that no triangle is possible. The following 
work verifies this fact. 
To find B, use the relationship 

sin B sin A 



obtaining 



sin B = 



b a 

720 sin 47°23' 



230 



a -230 



The logarithmic work follows: 

log 720 = 2.8573 
log sin 47°23' = 9.8668 - 10 
12.7241 - 10 
log 230 = 2.3617 




Fig. 18-7. 



log sin B = 10.3624 - 10 
= 0.3624. 
Since log sin B > 0, sin B would have to be greater than 1, which is impossible. 
Therefore, there is no solution. 



EXERCISE 18-1 

In each of the problems from 1 to 12, solve the given triangle by the law of sines. 

1. a = 12.30, A = 36°25', B = 44°37'. 

2. b = 12.18, A = 47°33', B = 67°51'. 

3. c = 461.3, B = 67°19', C = 23°14'. 

4. b = 0.6384, B = 39°39', C = 87°16'. 

5. a = 6.714, A = 37°53', C = 136°36'. 

6. c = 7832, A = 68°39 ; , B = 43°58 ; . 

7. a = 21.23, c = 64.21, C = 62°31'. 

8. b = 0.8146, c = 31.63, B = 11°10 / . 

9. a = 987.4, b = 503.6, A = 54°13'. 

10. a = 0.003862, c = 0.0008157, A = 26 13 ; . 

11. b = 1.386, c = 2.451, 5 = 83°19 ; . 

12. b = 4.395, c = 9.806, C = 37°46'. 

13. A surveyor wishes to find the distance across a stream from point A to point B. 
He finds that the distance from A to a point C on the same side of the stream is 
687.4 feet, and angles BAC and BCA are 49°53' and &8°16 ; , respectively. Find 
the distance AB. 

14. A surveyor was running a line due west when he reached a swamp. From the 
edge of the swamp he ran a line S 63° W for 2500 feet, and from this point he 
ran a line N 27 23' W. How far had he gone on this line when he reached his 
original line produced? How far was it across the swamp? 

15. A building 63.7 feet high stands on the top of a hill. From a point at the foot 
of the hill the angles of elevation to the top and bottom of the building are 
42°16' and 38°31', respectively. Find the height of the hill. 



296 



Solution of the General Triangle 



Sec. 18-4 



16. From a certain point on the ground the angle of elevation to the top of a building 
is 46° 17'. From a point on the ground 83 feet nearer the building the angle of 
elevation is 68°23'. Assuming that the ground is level, find the height of the 
building. 

17. One side and a diagonal of a parallelogram are 14.63 inches and 21.4 inches, 
respectively. The angle between the diagonals and opposite the given side is 
116°23'. Find the length of the other diagonal. 

18. It is necessary to measure the distance between two artillery pieces A and B. 
The angle of depression from an observation point C to gun A is 24°47'. Sound 
travels at the rate of 1 140 feet per second, and the sounds from guns A and B 
reach C in 2.3 and 1.7 seconds, respectively. Find the distance AB } assuming 
that points A } B ) and C lie in the same plane. 

19. A body is acted on by two forces, Fi = 2643 pounds and F 2 = 2341 pounds. 
The resultant F 3 lies on a line making an angle of 46°33' with F x . Find F 3 and 
the angle between the lines of action of Fi and F 2 . (The resultant of two 
forces is their vector sum.) 

20. A buoy, located at a point B, is 6 miles from a point A at one end of an island 
and 10 miles from a point C at the other end of the island. If the angle BAC is 
132°16', find the distance between the points A and C on the island. 

18-5. THE LAW OF COSINES 

Law of Cosines* Let ABC be any triangle. Then 
(18-8) a 2 = b 2 + c 2 - 2bc cos A, 

(18-9) b 2 = a 2 + c 2 - 2ac cos B, 

(18-10) c 2 = a 2 + b 2 - 2ab cos C. 

These relationships constitute the law of cosines. 





< 


n 








V 


h 


>^o 




A ^ 


X 


D 


c-x 




V- 




- c 







(a) 



B 



Fig. 18-8. 




Proof. We shall establish (18-8) by considering the case when A 
is acute, as in Fig. 18-8 (a) and the case when A is obtuse as in 
Fig. 18-8 (&). The case A = 90° involves no difficulty, and it will 
therefore be omitted. 

Let h denote the altitude from C to the side AB. Also, let x 
denote the length AD. Hence, DB is c — x in Fig. 18-8 (a) and is 
c + x in Fig. 18-8(6). 



Sec. 1 8-6 Solution of the General Triangle 297 

In Fig. 18-8 (a), 

(c - x) 2 + h 2 = a 2 , 
and 

X 2 + h 2 = b 2 . 
Subtracting, we have c 2 — 2cx = a 2 — 6 2 , or 

a 2 = b 2 + c 2 - 2cz. 



a 2 = 6 2 + c 2 — 26c cos A. 
(c + a;) 2 + h 2 = a 2 , 



Since # = b cos A, 
In Fig. 18-8(6), 

and 

x 2 + h 2 = 6 2 . 

Subtracting, we find that c 2 + 2c# = a 2 — 6 2 , or 

a 2 = 6 2 + c 2 + 2cz. 

Since x = — 6 cos A in this case, 

a 2 = b 2 + c 2 — 26c cos A. 

By drawing altitudes to the other sides and proceeding in a 
similar manner, we obtain (18-9) and (18-10). 

Note. For a simple algebraic proof of the law of cosines, see prob- 
lem 22 in Exercise 18-4. The law of cosines applies equally well if 
ABC is a right triangle. In this case, one of the formulas reduces 
to the pythogorean theorem, since cos 90° = 0. 

18-6. SOLUTION OF CASE III AND CASE IV BY THE LAW OF COSINES 

Since the law of cosines is expressed by formulas involving addi- 
tion and subtraction, it is not well adapted to logarithmic computa- 
tion and its use is not recommended unless the given sides are 
easily squared. 

Example 18-6. Given b = 9.0, c = 13.0, A = 115°10', solve the triangle ABC. 

Solution: By the law of cosines, 

a 2 = b 2 + c 2 - 26c cos A 

= 9 2 + (13) 2 - 2(9) (13) cos 115°10' 
= 81 + 169 - 234( - cos 64°50') 
= 250 + 234 (0.4253) = 349.52. 
Hence, 

a = 18.7. 
We employ the law of sines to find angle B. Thus, 

sin£ ^ sin 115°10' sin 64°50 / 
9 18.7 18.7 

and B = 25°50'. 

Therefore, C = 180° - (115°10' + 25°50') = 39°. 



298 Solution of the General Triangle Sec. 1 8-6 

Example 18-7. Given a = 3, 6 = 5, c = 7. Find the angles. 

Solution: From a 2 = 6 2 + c 2 — 26c cos ^4, we have the formula 

6 2 + c 2 - a 2 

COS A = -^ 

Therefore, _ ^ 

Hence, A = 21*47'. 

Similarly, B = 38°13' and C = 120°. 
We can check these by the equation A -f 5 -f C = 180°. 

EXERCISE 18-2 

In each of the problems from 1 to 6, solve the given triangle by the law of cosines. 
1. a = 300, 6 = 250, C - 58°40'. 2. a = 50, c = 240, 5 = 110°50'. 

3. 6 = 65, c = 310, A = 67°10 / . 4. a = 130, c = 90, £ = 100°20'. 

5. 6 = 50, c = 110, A = 150°. 6. a = 1.63, 6 = 3.45, C = 26°10'. 

7. If a = 15, b = 12, c - 20, find A. 

8. If a = 25, 6 = 30, and c = 35, find B. 

9. If a = 100, 6 = 300, c = 500, find C. 

10. If a = 15, 6 = 12, and c = 20, find B. 

11. If a = 16, 6 = 17, and c = 18, find A, B y C. 

12. If a = 260, 6 = 322, c = 481, find A, £, C. 

13. The distance between two points A and B cannot be measured directly. 
Accordingly, a third point C is selected, and it is found that AC = 3000 feet, 
£C = 4500 feet, and angle ACB = 46°20'. Find the distance A£. 

14. Two sides of a parallelogram are 125 feet and 200 feet, and the included angle is 
110°30'. Find the length of the longer diagonal of the parallelogram and also 
the angle between that diagonal and a longer side of the parallelogram. 

15. Two sides of a triangular plot of ground are 250 feet and 200 feet, and the 
included angle is 67°33'. Find the perimeter of the plot. 

16. Two sides of a parallelogram are 700 feet and 420 feet, and one diagonal is 
600 feet. Find the length of the other diagonal. 

17. In a triangle ABC, a = 25, b = 27, and the median from A is 20. Find c, A, B, C. 

18-7. THE LAW OF TANGENTS 

Law of Tangents. Let ABC be any triangle. Then the following 
relationships exist between two sides and the angles opposite them : 

. tan i (A - B) 

(18-n) ^z^ = 2 :, 

a + b tan - (A + B) 



Sec. 18-7 Solution of the General Triangle 299 

(18-12) 



6-c ta 4< B ~ C > 



(18-13) 



b + c tan ^ (B + C) 

tan J (C - A) 

c — a 2 ' 



C + a tan±(C + A) 



These relationships constitute the law of tangents. 

Proof. Let us denote the common ratio of the law of sines by r. 
Thus, a = r sin A, 6 = r sin J5, c = r sin C Then 

a — b __ r sin A — r sin B _ sin A — sin J? 
a + 6 ~~ r sin A + r sin B "" sin A + sin B 

Substituting from (7-27) and (7-28) of Section 7-4, we have 

smA+sinB 2 sin * (A + fi) cos 1 (4 _ £) 

Hence, 

. tan I (A - B) 

(18-11) ^l = — T^ "• 

a + 6 tani(A + B) 

A similar procedure may be followed to prove (18-12) and 
(18-13). 

We shall now use the law of tangents to solve a triangle in which 
two sides and the included angle are given. Note that this law is 
well adapted to logarithmic computation. 

Example 18-8. Given b = 249, c = 372, A = 56°22', solve the triangle ABC. 
Solution: To find C — B, we use the formula 

tan|(C-*)=^|tan|((7+B). 

Here c - b = 123, c + b = 621, C + B = 180° - A = 123°38', | (C + B) = 61°49'. 
Therefore, 19 „ 

tan±(C-£)=±gtan61°49', 
and - 

log tan i (C - B) = log 123 + log tan 61°49' - log 621. 



300 Solution of f/ia Genera/ Triangle Sec. 1 8-7 

The logarithmic work follows: 

log 123 = 2.0899 
log tan 61°49' = 0.2710 

12.3609 - 10 
log 621 = 2.7931 

log tan i (O - B) = 9.5678 - 10 



Hence, 
and 

Check: 



i (C - B) = 20°17'. 
C=g(C+B)+|(C-B) = 82°6', 
£ = J (C 4- B) - ~ (C - B) = 41°32'. 



A +B +C = 56°22' + 82°6' + 41°32' = 180° 
To find a, we use the law of sines. Thus, 

249sin56°22' 010 
a== sin41°32' =313 - 



18-8. THE HALF-ANGLE FORMULAS 

The following relationships are very convenient for the loga- 
rithmic solution of Case IV, where the three sides a, b, and c are 
known : 



(18-14) tan 2 = V s(s - a) » 

(18-15) tan 2 = y g(s - b) ' 

(18-16) ta C = ^( i -aH7^6) < 

7 2 r s(s — c) 

In these formulas, 5 = - (a + b + c). 

Proof. From (7-16) in Section 7-3, we have 

, 9 A 1 — cos A 

tan J 75- = 7—: j • 

2 1 + cos A 

Also, from the law of cosines, 

- b 2 + c 2 - a 2 

COS it = ?TT • 

26c 
Therefore 

' , b 2 + c 2 - a 2 a 2 -(b-c) 2 (a + b - c) (a -b + c) 

1 — cos A = 1 ^r = si = sr ' 

26c 26c 26c 

and 

1lMi _ , , 6 2 + c 2 - o 2 _ (6 + c) 2 - a 2 _ (b + c + a) (b + c - a) , 
l + oosil = l + ^ Wc Wc 



Sec. 18-8 Solution of the General Trhngle 301 

If we let a + 6 + c = 2s, then 

a + b - c = a + b + c — 2c = 2(s — c), 
a - 6 + c = a + 6 + c - 26 = 2(s - 6), 

6 + c — a = a + 6 + c — 2a = 2(s — a). 

Therefore, 

f o n 2 A - (<* + & - c) (a -- 6 + c) _ (s - c) (s - 6) 
™ n 2 " (6 + c + a) (6 + c - a) " s(s - a) 

and 



(18-14) tanA = 4/IiZSZS. 

2 r s(s — a) 

We can derive (18-15) and (18-16) in a similar manner. 

Example 18-9. Given: a = 379, b = 227, c = 416, find the angles of the triangle. 

Solution: Here 2s=a+b+c = 1022. Then 

8 =511, 
s - a = 132, 
s - 6 = 284, 
s - c = 95. 
Hence, 

. A j / (284) (9 5) 
tan 2- = T ( 511)(132) ' 
The calculations by logarithms follow: 



log 284 = 2.4533 log 511 = 2.7084 

log 95 = 1.9777 log 132 = 2.1206 

log (284) (95) = 24.4310 - 20 4.8290 
log (511) (132) = 4.8290 

2 1 19.6020 - 20 



log tan y = 9.8010 - 10 

a . ., - A = 64°38'. 

Similarly, 



tanf = |/p»- 
tan 



2 ~ V (511) (284) ' 



2 ~ K ( 



2 f (511) (95) 
Hence, we find that B = 32°46' and C = 82°36\ 
Check: 

A+B +C = 64°38' + 32°46' + 82°36' = 180°. 

EXERCISE 18-3 

In each of the problems from 1 to 10, solve the given triangle by the law of 
tangents if an angle is given, or by the half-angle formulas if three sides are given 
1. a = 50, 6 = 60, C = 60°. * 2. 6 = 17.1, c = 22.3, A = 21°16'. 

3. o = 230, c = 106, B = 95°10'. 4. b = 79.3, c = 113, A = 133°14'1 



302 Solution of the General Triangle Sec. 1 8-8 

5. 6 = 41.82, c = 75.89, A = 78°49'. 6. a = 0.1028, 6 = 0.8726, C = 148°13\ 
7. a = 625, 6 = 725, c = 825. 8. a = 60.65, 6 = 38.64, c = 23.57. 

9. a = 67450, b = 84380, c = 98630. 10. a = 0.1146, 6 = 0.3184, c = 0.6379. 

11. The diagonals of a parallelogram are 6 inches and 10 inches, and they intersect 
at an angle of 63°. Find the sides of the parallelogram. 

12. Points A and B are separated by an obstacle. In order to find the distance 
between them, a third point C is selected and it is found that AC = 126 rods 
and BC = 185 rods. The angle subtended at C by AB is 96°14'. Find AB. 

13. Two circles whose radii are 14 and 17 inches respectively intersect. The angle 
between the tangents to the circles at either point of intersection is 38°46'. 
Find the distance between the centers of the circles. 

14. The sides of a parallelogram are 13.4 inches and 18.5 inches, and one diagonal 
is 15.6 inches. Find the angles and the other diagonal of the parallelogram. 

15. Three circles whose radii are 10, 11, and 12 inches, respectively, are tangent to 
each other externally. Find the angles of the triangle formed by joining their 
centers. 

16. The sides of a triangular field are in the proportion 4:5:6. The area of the field is 
18 acres. If there are 160 square rods in an acre, find the length of each side 
of the field in rods. 

17. In triangle ABC y prove the following: 



c 1 n 

cos-C 

a+b C0S l (A ~ B) 



c , 1 n 

sin -C 

These formulas are called Mollweide's equations. They may be used in checking 
the solution of a triangle. 

18-9. AREA OF A TRIANGLE 

We can readily see that the area K of the triangle in Fig. 18-1 is 

(18-17) K = ^ch =^cb sin A. 

In either triangle, h = b sin A. In like manner, we obtain 

(18-18) K = lac sin B, 



(18-19) K = \ab sin C. 



2 
1 

2° 



c sin B 

By substituting ^ for 6 from the law of sines, we may 

sin C 

transform (18-17) to obtain 



Sec. 1 8-9 Solution of the General Triangle 303 

By cyclic interchanges of letters, we obtain 
(18-21) K = a2 sin B sinC 



(18-22) K = 



2 sin A 
b 2 sin A sin C 



2 sin B 

To derive a formula for finding the area of a triangle when its 
three sides are given, we first transform (18-17) in the following 
manner: 



K = ybc sin A = ^bc sin f 2 • -^-j 
2 6c ( 2sin 2~ COS 2~) 



2 

, . A A 
= 6c sm y cos y 

But, from (7-14) and (7-15) in Section 7-3, 



sin 



A A /l — cos A , A A f\ 

- = j/ and cos y =4/- 



2 

Using the values from Section 18-8 for 1 — cos A and 1 + cos A, we 
have 



. A A /(s — c) (s — b) , A A /s(s — a) 

sm 2=V — Fc — - and C0S 2 = T be— 

Consequently, the formula for area in terms of the sides is 



(18-21) K = be |/ (S - C) 6 f ~ 6) V^HS 

= \/s(s — a) (s — b) (s — c). 
The following examples illustrate the use of the area formulas. 

Example 18-10. Find the area of the triangle in Example 18-8, in which b ='249, 
c = 372, A = 56°22'. 

Solution: Since two sides and the included angle are given, (18-17) may be used* 
Thus, 

K = he sin A = ^(249) (372) sin 56°22>. 

The logarithmic work follows : 

log 0.5 = 9.6990 - 10 
log 249 = 2.3962 
log 372 = 2.5705 
log sin 56°22' = 9.9205 - 10 
log K = 24.5862 - 20 
K = 38,600. 



304 Solution of the General Triangle Sec. 1 8-9 

Example 18-11. Find the area of the triangle in Example 18-1, in which A 
= 38°14', B = 67°20', c = 329. 

Solution: Since two angles and a side are given, (18-20) may be used. In this case, 
„ c* sin A sin B _ (329)* sin 38°14' sin 67°20' _ QO inn 
* - 2 sinC ~ 2ih774 5 26 7 ~ 32 ' 100 ' 



Example 18-12. Find the area of the triangle in Example 18-9, in which a = 379, 
b = 227, c = 416. 

Solution: In this solution, (18-21) is used. We have 



K = V*(* -«)(*- &) (* - c) = V(5H) (132) (284) (95) = 42,700. 



EXERCISE 18-4 

In each of the problems from 1 to 8, find the area of the given triangle. 
1. a = 12.30, A = 36°25', B = 44°37'. 2. c = 461.3, B = 67°19 ; , C = 23°14'. 
3. a = 987.4, 6 = 503.6, A = 54°13'. 4. 6 = 4.395, c = 9.806, C = 37°46'. 

5. 6 = 65, c = 310, A = 67°10 ; . 6. a = 300, 6 = 250, C = 58°40'. 

7. a = 15, 6 = 12, c = 20. 8. a = 100, 6 = 300, c = 500. 

9. In triangle ABC, let r be the radius of the inscribed circle. Prove that K = rs 



and, therefore, that r 



= yi/ fr ~ a ) ( s ~ 6) ( s ""g) 



10. Find the radius of the circle inscribed in the triangle whose sides are 48.92 
feet, 63.86 feet, and 72.31 feet. 

11. A cylindrical tank is to be built on a triangular lot having sides whose lengths 
are 200 feet, 186 feet, and 176 feet. Find the radius of the largest such tank 
which can be built on the lot. 

12. In triangle ABC, let R be the radius of the circumscribed circle. Show that 

2R= « = » = « . 

sin A sin B sin C 

13. In triangle ABC, show that R = -rr? t where R is the radius of the circum- 
scribed circle and K is the area of the triangle. 

14* The sides of a triangle are 23, 29, and 46 feet. Find the areas of the triangle 
and the inscribed and circumscribed circles. 

15. The sides of a triangular plot of grass are 42 feet, 65 feet, and 87 feet. Find the 
minimum radius of action of an automatic lawn sprinkler which will water all 
parts of the plot from the same point. 

16. An arc of a circle of radius r subtends a central angle 6. Show that the area 

bounded by this arc and its chord is jr r 2 (6 — sin 6). 

17. Find the area of the largest pentagon which can be cut from a circular piece of 
metal 4 feet in radius. How much metal is wasted? 

18. In triangle ABC, prove that the median from any vertex to the side opposite 
divides the angle at that vertex into two parts whose sines are proportional to 
the lengths of the parts into which the side opposite is divided by the median. 



Sec. 18-9 Solution of the General Triangle 305 

19. In triangle ABC, prove that 

cos A , cos B , cos C a 2 + 6 2 + c 2 
a b c 2abc 

20. In triangle ABC, prove that 

a + b + c = (b + c) cos A + (c + a) cos B + (a + 6) cos C 

21. In triangle ABC, prove that 

a 2 + b 2 + c 2 = a 2 (cos 2 C + sin 2 B) + 6 2 (cos 2 A + sin 2 C) + c 2 (cos 2 B + sin 2 A). 

22. In triangle ABC, show that 

a = 6 cos C + c cos 5, 

6 = c cos A + a cos C, 

c = a cos B -f 6 cos A. 
Multiply the first equation by a, the second by b, and the third by c, to give a 
second proof of the law of cosines; that is, prove (18-8) by showing that 
b 2 + c 2 - a 2 = 26c cos A. Similarly prove (18-9) and (18-10). 

23. Consider any triangle ABC. If a > b, prove that A > B. If A > B, prove that 
a > b. 

24. In triangles ABC and A'B'C, let A and A 1 , B and B', C and C be pairs of 
corresponding vertices, and let the corresponding sides be a and a', b and V, 
c and c'. If a = a', 6 = 6', and C > C", prove that c > c r . If a = a', 6 = &', 
and c > c', prove that C > C". 



Appendix A 



30* 



TABLE I 
Four-Place Values of Functions op Numbers 



t 


sin t 


cos t 


tan t 


cot t 


sec t 


esc t 


.00 


.0000 


1.0000 


.0000 




1.000 




.01 


.0100 


1.0000 


.0100 


99.997 


1.000 


100.66 


.02 


.0200 


.9998 


.0200 


49.993 


1.000 


50.00 


.03 


.0300 


.9996 


.0300 


33.323 


1.000 


33.34 


.04 


.0400 


.9992 


.0400 


24.987 


1.001 


25.01 


.05 


.0500 


.9988 


.0500 


19.983 


1.001 


20.01 


.06 


.0600 


.9982 


.0601 


16.647 


1.002 


16.68 


.07 


.0699 


.9976 


.0701 


14.262 


1.002 


14.30 


.08 


.0799 


.9968 


.0802 


12.473 


1.003 


12.51 


.09 


.0899 


.9960 


.0902 


11.081 


1.004 


11.13 


.10 


.0998 


.9950 


.1003 


9.967 


1.005 


10.02 


.11 


.1098 


.9940 


.1104 


9.054 


1.006 


9.109 


.12 


.1197 


.9928 


.1206 


8.293 


1.007 


8.353 


.13 


.1296 


.9916 


.1307 


7.649 


1.009 


7.714 


.14 


.1395 


.9902 


.1409 


7.096 


1.010 


7.166 


.15 


.1494 


.9888 


.1511 


6.617 


1.011 


6.692 


.16 


.1593 


.9872 


.1614 


6.197 


1.013 


6.277 


.17 


.1692 


.9856 


.1717 


5.826 


1.015 


5.911 


.18 


.1790 


.9838 


.1820 


5.495 


1.016 


5.586 


.19 


.1889 


.9820 


.1923 


5.200 


1.018 


5.295 


.20 


.1987 


.9801 


.2027 


4.933 


1.020 


5.033 


.21 


.2085 


.9780 


.2131 


4.692 


1.022 


4.797 


.22 


.2182 


.9759 


.2236 


4.472 


1.025 


4.582 


.23 


.2280 


.9737 


.2341 


4.271 


1.027 


4.386 


.24 


.2377 


.9713 


.2447 


4.086 


1.030 


4.207 


.25 


.2474 


.9689 


.2553 


3.916 


1.032 


4.042 


.26 


.2571 


.9664 


.2660 


3.759 


1.035 


3.890 


.27 


.2667 


.9638 


.2768 


3.613 


1.038 


3.749 


.28 


.2764 


.9611 


.2876 


3.478 


1.041 


3.619 


.29 


.2860 


.9582 


.2984 


3.351 


1.044 


3.497 


.30 


.2955 


.9553 


.3093 


3.233 


1.047 


3.384 


.31 


.3051 


.9523 


.3203 


3.122 


1.050 


3.278 


.32 


.3146 


.9492 


.3314 


3.018 


1.053 


3.179 


.33 


.3240 


.9460 


.3425 


2.920 


1.057 


3.086 


.34 


.3335 


.9428 


.3537 


2.827 


1.061 


2.999 


.35 


.3429 


.9394 


.3650 


2.740 


1.065 


2.916 


.36 


.3523 


.9359 


.3764 


2.657 


1.068 


2.839 


.37 


.3616 


.9323 


.3879 


2.578 


1.073 


2.765 


.38 


.3709 


.9287 


.3994 


2.504 


1.077 


2.696 


.39 


.3802 


.9249 


.4111 


2.433 


1.081 


2.630 


t 


sin t 


cos t 


tan t 


cot t 


' sec t 


CSC t i. 



310 



Appendix A 



TABLE I (continued) 



t 


sin t 


cos t 


tan t 


cot t 


sec t 


esc t 


.40 


.3894 


.9211 


.4228 


2.365 


1.086 


2.568 


.41 


.3986 


.9171 


.4346 


2.301 


1.090 


2.509 


.42 


.4078 


.9131 


.4466 


2.239 


1.095 


2.452 


.43 


.4169 


.9090 


.4586 


2.180 


1.100 


2.399 


.44 


.4259 


.9048 


.4708 


2.124 


1.105 


2.348 


.45 


.4350 


.9004 


.4831 


2.070 


1.111 


2.299 


.46 


.4439 


.8961 


.4954 


2.018 


1.116 


2.253 


.47 


.4529 


.8916 


.5080 


1.969 


1.122 


2.208 


.48 


.4618 


.8870 


.5206 


1.921 


1.127 


2.166 


.49 


.4706 


.8823 


.5334 


1.875 


1.133 


2.125 


.50 


.4794 


.8776 


.5463 


1.830 


1.139 


2.086 


.51 


.4882 


.8727 


.5594 


1.788 


1.146 


2.048 


.52 


.4969 


.8678 


.5726 


1.747 


1.152 


2.013 


.53 


.5055 


.8628 


.5859 


1.707 


1.159 


1.978 


.54 


.5141 


.8577 


.5994 


1.668 


1.166 


1.945 


.55 


.5227 


.8525 


.6131 


1.631 


1.173 


1.913 


.56 


.5312 


.8473 


.6269 


1.595 


1.180 


1.883 


.57 


.5396 


.8419 


.6410 


1.560 


1.188 


1.853 


.58 


.5480 


.8365 


.6552 


1.526 


1.196 


1.825 


.59 


.5564 


.8309 


.6696 


1.494 


1.203 


1.797 


.60 


.5646 


.8253 


.6841 


1.462 


1.212 


1.771 


.61 


.5729 


.8196 


.6989 


1.431 


1.220 


1.746 


.62 


.5810 


.8139 


.7139 


1.401 


1.229 


1.721 


.63 


.5891 


.8080 


.7291 


1.372 


1.238 


1.697 


.64 


.5972 


.8021 


.7445 


1.343 


1.247 


1.674 


.65 


.6052 


.7961 


.7602 


1.315 


1.256 


1.652 


.66 


.6131 


.7900 


.7761 


1.288 


1.266 


1.631 


.67 


.6210 


.7838 


.7923 


1.262 


1.276 


1.610 


.68 


.6288 


.7776 


.8087 


1.237 


1.286 


1.590 


.69 


.6365 


.7712 


.8253 


1.212 


1.297 


1.571 


.70 


.6442 


.7648 


.8423 


1.187 


1.307 


1.552 


.71 


.6518 


.7584 


.8595 


1.163 


1.319 


1.534 


.72 


.6594 


.7518 


.8771 


1.140 


1.330 


1.517 


.73 


.6669 


.7452 


.8949 


1.117 


1.342 


1.500 


.74 


.6743 


.7385 


.9131 


1.095 


1.354 


1.483 


.75 


.6816 


.7317 


.9316 


1.073 


1.367 


1.467 


.76 


.6889 


.7248 


.9505 


1.052 


1.380 


1.452 


.77 


.6961 


.7179 


.9697 


1.031 


1.393 


1.437 


.78 


.7033 


.7109 


.9893 


1.011 


1.407 


1.422 


.79 


.7104 


.7038 


1.009 


.9908 


1.421 


1.408 


t 


sint 


cos t 


tan t 


cot t 


sec t 


esc t 



Appendix A 



311 



TABLE I (continued) 



t 


sin t 


cos t 


tan t 


cot t 


sec t 


CSC t 


.80 


.7174 


.6967 


1.030 


.9712 


1.435 


1.394 


.81 


.7243 


.6895 


1.050 


.9520 


1.450 


1.381 


.82 


.7311 


.6822 


1.072 


.9331 


1.466 


1.368 


.83 


.7379 


.6749 


1.093 


.9146 


1.482 


1.355 


.84 


.7446 


.6675 


1.116 


.8964 


1.498 


1.343 


.85 


.7513 


.6600 


1.138 


.8785 


1.515 


1.331 


.86 


.7578 


.6524 


1.162 


.8609 


1.533 


1.320 


.87 


.7643 


.6448 


1.185 


.8437 


1.551 


1.308 


.88 


.7707 


.6372 


1.210 


.8267 


1.569 


1.297 


.89 


.7771 


.6294 


1.235 


.8100 


1.589 


1.287 


.90 


.7833 


.6216 


1.260 


.7936 


1.609 


1.277 


.91 


.7895 


.6137 


1.286 


.7774 


1.629 


1.267 


.92 


.7956 


.6058 


1.313 


.7615 


1.651 


1.257 


.93 


.8016 


.5978 


1.341 


.7458 


1.673 


1.247 


.94 


.8076 


.5898 


1.369 


.7303 


1.696 


1.238 


.95 


.8134 


.5817 


1.398 


.7151 


1.719 


1.229 


.96 


.8192 


.5735 


1.428 


.7001 


1.744 


1.221 


.97 


.8249 


.5653 


1.459 


.6853 


1.769 


1.212 


.98 


.8305 


.5570 


1.491 


.6707 


1.795 


1.204 


.99 


.8360 


.5487 


1.524 


.6563 


1.823 


1.196 


1.00 


.8415 


.5403 


1.557 


.6421 


1.851 


1.188 


1.01 


.8468 


.5319 


1.592 


.6281 


1.880 


1.181 


1.02 


.8521 


.5234 


1.628 


.6142 


1.911 


1.174 


1.03 


.8573 


.5148 


1.665 


.6005 


1.942 


1.166 


1.04 


.8624 


.5062 


1.704 


.5870 


1.975 


1.160 


1.05 


.8674 


.4976 


1.743 


.5736 


2.010 


1.153 


1.06 


.8724 


.4889 


1.784 


.5604 


2.046 


1.146 


1.07 


.8772 


.4801 


1.827 


.5473 


2.083 


1.140 


1.08 


.8820 


.4713 


1.871 


.5344 


2.122 


1.134 


1.09 


.8866 


.4625 


1.917 


.5216 


2.162 


1.128 


1.10 


.8912 


.4536 


1.965 


.5090 


2.205 


1.122 


1.11 


.8957 


.4447 


2.014 


.4964 


2.249 


1.116 


1.12 


.9001 


.4357 


2.066 


.4840 


2.295 


1.111 


1.13 


.9044 


.4267 


2.120 


.4718 


2.344 


1.106 


1.14 


.9086 


.4176 


2.176 


.4596 


2.395 


1.101 


1.15 


.9128 


.4085 


2.234 


.4475 


2.448 


1.096 


1.16 


.9168 


.3993 


2.296 


.4356 


2.504 


1.091 


1.17 


.9208 


.3902 


2.360 


.4237 


2.563 


1.086 


1.18 


.9246 


.3809 


2.427 


.4120 


2.625 


1.082 


1.19 


.9284 


.3717 


2.498 


.4003 


2.691 


1.077 


t 


sin t 


cos t 


tan t 


cot t 


sec t 


CSC t 



nt 



Appendix A 



TABLE I (continued) 



t 


sin t 


cos J 


tan t 


cot t 


sec t 


esc t 


1.20 


.9320 


.3624 


2.572 


.3888 


2.760 


1.073 


1.21 


.9356 


.3530 


2.650 


.3773 


2.833 


1.069 


1.22 


.9391 


.3436 


2.733 


.3659 


2.910 


1.065 


1.23 


.9425 


.3342 


2.820 


.3546 


2.992 


1.061 


1.24 


.9458 


.3248 


2.912 


.3434 


3.079 


1.057 


1.25 


.9490 


.3153 


3.010 


.3323 


3.171 


1.054 


1.26 


.9521 


.3058 


3.113 


.3212 


3.270 


1.050 


1.27 


.9551 


.2963 


3.224 


.3102 


3.375 


1.047 


1.28 


.9580 


.2867 


2.341 


.2993 


3.488 


1.044 


1.29 


.9608 


.2771 


3.467 


.2884 


3.609 


1.041 


1.30 


.9636 


.2675 


3.602 


.2776 


3.738 


1.038 


1.31 


.9662 


.2579 


3.747 


.2669 


3.878 


1.035 


1.32 


.9687 


.2482 


3.903 


.2562 


4.029 


1.032 


1.33 


.9711 


.2385 


4.072 


.2456 


4.193 


1.030 


1.34 


.9735 


.2288 


4.256 


.2350 


4.372 


1.027 


1.35 


.9757 


.2190 


4.455 


.2245 


4.566 


1.025 


1.36 


.9779 


.2092 


4.673 


.2140 


4.779 


1.023 


1.37 


.9799 


.1994 


4.913 


.2035 


5.014 


1.021 


1.38 


.9819 


.1896 


5.177 


.1931 


5.273 


1.018 


1.39 


.9837 


.1798 


5.471 


.1828 


5.561 


1.017 


1.40 


.9854 


.1700 


5.798 


.1725 


5.883 


1.015 


1.41 


.9871 


.1601 


6.165 


.1622 


6.246 


1.013 


1.42 


.9887 


.1502 


6.581 


.1519 


6.657 


1.011 


1.43 


.9901 


.1403 


7.055 


.1417 


7.126 


1.010 


1.44 


.9915 


.1304 


7.602 


.1315 


7.667 


1.009 


1.45 


.9927 


.1205 


8.238 


.1214 


8.299 


1.007 


1.46 


.9939 


.1106 


8.989 


.1113 


9.044 


1.006 


1.47 


.9949 


.1006 


9.887 


.1011 


9.938 


1.005 


1.48 


.9959 


.0907 


10.983 


.0910 


11.TJ29 


1.004 


1.49 


.9967 


.0807 


12.350 


.0810 


12.390 


1.003 


1.50 


.9975 


.0707 


14.101 


.0709 


14.137 


1.003 


1.51 


.9982 


.0608 


16.428 


.0609 


16.458 


1.002 


1.52 


.9987 


.0508 


19.670 


.0508 


19.695 


1.001 


1.53 


.9992 


.0408 


24.498 


.0408 


24.519 


1.001 


1.54 


.9995 


.0308 


32.461 


.0308 


32.476 


1.000 


1.55 


.9998 


.0208 


48.078 


.0208 


48.089 


1.000 


1.56 


.9999 


.0108 


92.620 


.0108 


92.626 


1.000 


1.57 


1.0000 


.0008 


1255.8 


.0008 


1255.8 


1.000 


1.58 


1.0000 


-.0092 


. -108.65 


-.0092 


-108.65 


1.000 


1.59 


.9998 


-.0192 


-52.067 


-.0192 


-52.08 


1.000 


1.60 


. .9996 


-.0292 


-34.233 


-.0292 


-34.25 


1.000 


t 


sin t 


cost 


tan t 


cot t 


sec t 


esc t 



Appendix A 



313 



TABLE II 
Four-Place Values of Functions 





Sin 


Cos 


Tan 


Cot 


Sec 


Gsc 






0°00' 

10' 


.0000 


1.000 


.0000 




1.000 




90° 00' 

89° 50' 


029 


000 


029 


343.8 


000 


343.8 


20' 


058 


000 


058 


171.9 


000 


171.9 


40' 


30' 


.0087 


1.000 


.0087 


114.6 


1.000 


114.6 


30' 


40' 


116 


.9999 


116 


85.94 


000 


85.95 


20' 


0°50' 
lo 00' 

10' 


145 


999 


145 
.0175 


68.75 


000 


68.76 


10' 
89° 00* 

88° 50' 


.0175 


.9998 


57.29 


1.000 


57.30 


204 


998 


204 


49.10 


000 


49.11 


20' 


233 


997 


233 


42.96 


000 


42.98 


40' 


30' 


.0262 


.9997 


.0262 


38.19 


1.000 


38.20 


30' 


40' 


291 


996 


291 


34.37 


000 


34.38 


20' 


1°50' 
10' 


320 


995 


320 


31.24 


001 


31.26 


10' 
88° 00' 

87° 50' 


.0349 


.9994 
993 


.0349 

378 


28.64 


1.001 


28.65 


378 


26.43 


001 


26.45 


20' 


407 


992 


407 


24.54 


001 


24.56 


40' 


30' 


.0436 


.9990 


.0437 


22.90 


1.001 


22.93 


30' 


40' 


465 


989 


466 


21.47 


001 


21.49 


20' 


2° 50' 
3° 00' 

10' 


494 


988 


495 


20.21 


001 


20.23 


10' 
87° 00' 

86° 50' 


.0523 


.9986 


.0524 


19.08 
18.07 


1.001 


19.11 


552 


985 


553 


002 


18.10 


20' 


581 


983 


582 


17.17 


002 


17.20 


40' 


30' 


.0610 


.9981 


.0612 


16.35 


1.002 


16.38 


30' 


40' 


640 


980 


641 


15.60 


002 


15.64 


20' 


3° 50' 
4° 00' 

10' 


669 


978 


670 


14.92 


002 


14.96 


10' 
86° 00' 

85*50' 


.0698 


.9976 


.0699 


14.30 


1.002 


14.34 


727 


974 


729 


13.73 


003 


13.76 


20' 


756 


971 


758 


13.20 


003 


13.23 


40' 


30' 


.0785 


.9969 


.0787 


12.71 


1.003 


12.75 


30' 


40' 


814 


967 


816 


12.25 


003 


12.29 


20' 


4° 50' 
50 00' 

10' 


843 


964 


846 


11.83 


004 


11.87 


10' 
86° 00' 

84° 50* 


.0872 


.9962 


.0875 


11.43 


1.004 


11.47 


901 


959 


904 


11.06 


004 


11.10 


20' 


929 


957 


934 


10.71 


004 


10.76 


40' 


30' 


.0958 


.9954 


.0963 


10.39 


1.005 


10.43 


30' 


40' 


.0987 


951 


.0992 


10.08 


005 


10.13 


20? 


5° 50' 


.1016 


948 


.1022 


9.788 


005 


9.839 


10' 
84° W 


.1045 


.9945 


.1051 


9.514 


1.006 


9.567 


Cos 


Sin 


Cot 


Tan 


Csc 


Sec 







314 



Appendix A 







TABLE II 


(continued) 








Sin 


Cos 


Tan 


Cot 


Sec 


Csc 






e°oc 

ic 


.1045 


.9945 


.1051 


9.514 


1.006 


9.567 


84° 00' 

83° 50' 


074 


942 


080 


255 


006 


309 


2C 


103 


939 


110 


9.010 


006 


9.065 


40' 


30' 


.1132 


.9936 


.1139 


8.777 


1.006 


8.834 


30' 


40 


161 


932 


169 


556 


007 


614 


20' 


6° 50' 

roc 

10' 


190 


929 
.9925 


198 


345 


007 


405 


10' 
83° 00' 

82° 50' 


.1219 


.1228 


8.144 


1.008 


8.206 


248 


922 


257 


7.953 


008 


8.016 


20' 


276 


918 


287 


770 


008 


7.834 


40' 


30' 


.1305 


.9914 


.1317 


7.596 


1.009 


7.661 


30' 


40' 


334 


911 


346 


429 


009 


496 


20' 


7° 50' 
8°0C 

10' 


363 


907 


376 


269 


009 


337 


10' 
82° 00' 

81° 50' 


.1392 


.9903 


.1405 


7.115 


1.010 


7.185 
7.040 


421 


899 


435 


6.968 


010 


20' 


449 


894 


465 


827 


Oil 


6.900 


40' 


30' 


.1478 


.9890 


.1495 


6.691 


1.011 


6.765 


30' 


40' 


507 


886 


524 


561 


012 


636 


20' 


8° 50' 
9° 00' 

10' 


536 


881 


554 


435 


012 


512 


10' 
81° 00' 

80° 50' 


.1564 


.9877 
872 


.1584 


6.314 


1.012 


6.392 


593 


614 


197 


013 


277 


20' 


622 


868 


644 


6.084 


013 


166 


40' 


30' 


.1650 


.9863 


.1673 


5.976 


1.014 


6.059 


30' 


40' 


679 


858 


703 


871 


014 


5.955 


20' 


9° 50' 
10° 00' 

1C 


708 


853 


733 


769 


015 


855 


10' 
80° 00' 

79° 50' 


.1736 
765 


.9848 


.1763 


5.671 


1.015 


5.759 


843 


793 


576 


016 


665 


20' 


794 


838 


823 


485 


016 


575 


40' 


30' 


.1822 


.9833 


.1853 


5.396 


1.017 


5.487 


30' 


40' 


851 


827 


883 


309 


018 


403 


20' 


10° 50' 

11° oc 

10' 


880 


822 

.9816 

811 


914 


226 


018 


320 


10' 
79° 00' 

78° 50' 


.1908 


.1944 


5.145 


1.019 


5.241 


937 


.1974 


5.066 


019 


164 


20' 


965 


805 


.2004 


4.989 


020 


089 


40' 


30' 


.1994 


.9799 


.2035 


4.915 


1.020 


5.016 


30' 


40' 


.2022 


793 


065 


843 


021 


4.945 


20' 


11° 50' 
12° 00' 


051 


787 


095 


773 


022 


876 


10' 
78° 00' 


.2079 


.9781 


.2126 


4.705 


1.022 


4.810 




Cos 


Sin 


Cot 


Tan 


Csc 


Sec 


4 


■ 





Appendix A 



315 







TABLE I] 


'. (continued) 








Sin 


Cos 


Tan 


Cot 


Sec 


Csc 






ia° oo' 

10' 


.2079 


.9781 


.2126 


4.705 


1.022 


4.810 


78° 00' 

77° 50' 


108 


775 


156 


638 


023 


745 


20' 


136 


769 


186 


574 


024 


682 


40' 


30' 


.2164 


.9763 


.2217 


4.511 


1.024 


4.620 


30' 


40' 


193 


757 


247 


449 


025 


560 


20' 


12° 50' 

13° 00' 

10' 


221 


750 


278 


390 


026 


502 


10' 
77° 00' 

76° 50' 


.2250 


.9744 


.2309 


4.331 


1.026 


4.445 


278 


737 


339 


275 


027 


390 


20' 


306 


730 


370 


219 


028 


336 


40' 


30' 


.2334 


.9724 


.2401 


4.165 


1.028 


4.284 


30' 


40' 


363 


717 


432 


113 


029 


232 


20' 


13° 50' 

14° 00' 

10' 


391 


710 


462 
.2493 


061 


030 


182 


10' 
76° 00' 

75° 50' 


.2419 


.9703 


4.011 


1.031 


4.134 


447 


696 


524 


3.962 


031 


086 


20' 


476 


689 


555 


914 


032 


4.039 


40' 


30' 


.2504 


.9681 


.2586 


3.867 


1.033 


3.994 


30' 


40' 


532 


674 


617 


821 


034 


950 


20 7 


14° 50' 
15° 00' 

10' 


560 


667 


648 


776 


034 


906 


10' 
75° 00' 

74° 50' 


.2588 


.9659 


.2679 


3.732 


1.035 


3.864 


616 


652 


711 


689 


036 


822 


20' 


644 


644 


742 


647 


037 


782 


40' 


30' 


.2672 


.9636 


.2773 


3.606 


1.038 


3.742 


30' 


40' 


700 


628 


805 


566 


039 


703 


20' 


15° 50' 
16° 00' 

10' 


728 
.2756 

784 


621 


836 


526 


039 


665 


10' 
74° 00' 

73° 50' 


.9613 


.2867 


3.487 


1.040 


3.628 


605 


899 


450 


041 


592 


20' 


812 


596 


931 


412 


042 


556 


40' 


30' 


.2840 


.9588 


.2962 


3.376 


1.043 


3.521 


30' 


40' 


868 


580 


.2994 


340 


044 


487 


20' 


16° 50' 
17° 00' 

10' 


896 


572 


.3026 


305 


045 


453 


10' 
73° 00' 

72° 50' 


.2924 


.9563 


.3057 


3.271 


1.046 


3.420 


952 


555 


089 


237 


047 


388 


20' 


.2979 


546 


121 


204 


048 


356 


40' 


30' 


.3007 


.9537 


.3153 


3.172 


1.049 


3.326 


30' 


40' 


035 


528 


185 


140 


049 


295 


20' 


17° 50' 
18° 00' 


062 
.3090 


520 


217 


108 


050 


265 


10' 
72° W 


.9511 


.3249 


3.078 


1.051 


3.236 


Cos 


Sin 


Cot 


Tan 


Csc 


Sec 







316 



Appendix A 







TABLE II (continued) 








Sin 


Cos 


Tan 


Cot 


Sec 


Csc 






18° 00' 

10' 


.3090 


.9511 


.3249 


3.078 


1.051 
052 


3.236 


72*00' 

71* 50' 


118 


502 


281 


047 


207 


20' 


145 


492 


314 


3.018 


053 


179 


40' 


30' 


.3173 


.9483 


.3346 


2.989 


1.054 


3.152 


30' 


40' 


201 


474 


378 


960 


056 


124 


20' 


18*50' 

19*00' 

10' 


228 


465 


411 


932 


057 


098 


10' 
71* 00' 

70* 50' 


.3256 


.9455 


.3443 


2.904 


1.058 


3.072 


283 


446 


476 


877 


059 


046 


20' 


311 


436 


508 


850 


060 


3.021 


40' 


30' 


.3338 


.9426 


.3541 


2.824 


1.061 


2.996 


30' 


40' 


365 


417 


574 


798 


062 


971 


20' 


19* 50' 

20*00' 

10' 


393 


407 


607 
.3640 


773 


063 


947 


10' 
70° 00' 

69° 50' 


.3420 


.9397 


2.747 


1.064 
065 


2.924 
901 


448 


387 


673 


723 


20' 


475 


377 


706 


699 


066 


878 


40' 


30' 


.3502 


.9367 


.3739 


2.675 


1.068 


2.855 


30' 


40' 


529 


356 


772 


651 


069 


833 


20' 


20* 50' 

21* 00' 

10' 


557 


346 


805 


628 


070 


812 


10' 
69* 00' 

68° 50' 


.3584 


.9336 


.3839 


2.605 
583 


1.071 


2.790 


611 


325 


872 


072 


769 


20' 


638 


315 


906 


560 


074 


749 


40' 


30' 


.3665 


.9304 


.3939 


2.539 


1.075 


2.729 


30' 


40' 


692 


293 


.3973 


517 


076 


709 


20' 


21* 50' 

22*00' 

10' 


719 


283 


.4006 


496 


077 


689 


10' 
68* 00' 

67° 50' 


.3746 


.9272 


.4040 


2.475 


1.079 


2.669 


773 


261 


074 


455 


080 


650 


20' 


800 


250 


108 


434 


081 


632 


40' 


30' 


.3827 


.9239 


.4142 


2.414 


1.082 


2.613 


30' 


40' 


854 


228 


176 


394 


084 


595 


20' 


22* 50' 

23* 00' 

10' 


881 


216 


210 


375 


085 


577 


10' 
67* 00' 

66* 50' 


.3907 


.9205 


.4245 


2.356 


1.086 


2.559 


934 


194 


279 


337 


088 


542 


20' 


961 


182 


314 


318 


089 


525 


40' 


30' 


.3987 


.9171 


.4348 


2.300 


1.090 


2.508 


30' 


40' 


.4014 


159 


383 


282 


092 


491 


20' 


23* 50' 
24*00' 


041 


147 


417 


264 
2.246 


093 


475 


10' 
66* 00' 


.4067 


.9135 


.4452 


1.095 


2.459 




Cos 


Sin 


Cot 


Tan 


Csc 


Sec 




\ 



Appendix A 



317 



TABLE II (continued) 





Sin 


Cos 


Tan 


Cot 


Sec 


Csc 






24" 00' 

10' 


.4067 


.9135 


.4452 


2.246 


1.095 


2.459 


66" 00' 

65" 50' 


094 


124 


487 


229 


096 


443 


20' 


120 


112 


522 


211 


097 


427 


40' 


30' 


.4147 


.9100 


.4557 


2.194 


1.099 


2.411 


30' 


40' 


173 


088 


592 


177 


100 


396 


20' 


24" 50' 
25° 00' 

10' 


200 


075 


628 


161 


102 


381 


10' 
65" 00' 

64" 50' 


.4226 


.9063 


.4663 


2.145 


1.103 


2.366. 


253 


051 


699 


128 


105 


352 


20' 


279 


038 


734 


112 


106 


337 


40' 


30' 


.4305 


.9026 


.4770 


2.097 


1.108 


2.323 


30' 


40' 


331 


013 


806 


081 


109 


309 


20' 


25" 50' 
26" 00' 

10' 


358 


.9001 


841 


066 


111 


295 


10' 
64" 00' 

63" 50' 


.4384 


.8988 
975 


.4877 


2.050 


1.113 


2.281 


410 


913 


035 


114 


268 


20' 


436 


962 


950 


020 


116 


254 


40' 


30' 


.4462 


.8949 


.4986 


2.006 


1.117 


2.241 


30' 


40' 


488 


936 


.5022 


1.991 


119 


228 


20' 


26° 50' 

27° 00' 

10' 


514 


923 


059 
.5095 


977 


121 


215 


10' 
63" 00' 

62" 50' 


.4540 


.8910 


1.963 
949 


1.122 


2.203 


566 


897 


132 


124 


190 


20' 


592 


884 


169 


935 


126 


178 


40' 


30' 


.4617 


.8870 


.5206 


1.921 


1.127 


2.166 


30' 


40' 


643 


857 


243 


907 


129 


154 


20' 


27" 50' 

28" 00' 

10' 


669 


843 


280 


894 


131 


142 


10' 
62" 00' 

61" 50' 


.4695 


.8829 


.5317 


1.881 


1.133 


2.130 


720 


816 


354 


868 


134 


118 


20' 


746 


802 


392 


855 


136 


107 


40' 


30' 


.4772 


.8788 


.5430 


1.842 


1.138 


2.096 


30' 


40' 


797 


774 


467 


829 


140 


085 


20' 


28" 50' 
29" 00' 

10' 


823 


760 


505 


816 


142 


074 


10' 
61" 00' 

60" 50' 


.4848 


.8746 


.5543 
581 


1.804 


1.143 


2.063 


874 


732 


792 


145 


052 


20' 


899 


718 


619 


780 


147 


041 


40' 


30' 


.4924 


.8704 


.5658 


1.767 


1.149 


2.031 


30' 


40' 


950 


689 


696 


756 


151 


020 


20' 


29" 50' 
30" 00' 


.4975 


675 


735 


744 


153 


010 


10' 
60" 00' 


.5000 


.8660 


.5774 


1.732 


1.155 


2.000 


Cos 


Sin 


Cot 


Tan 


Csc 


Sec 







318 



Appendix A 







TABLE II {continued) 






V 


Sin 


Cos 


Tan 


Cot 


Sec 


Csc 






30° 00' 

10' 


.5000 


.8660 
646 


.5774 


1.732 


1.155 
157 


2.000 


60° 00' 

59° 50' 


025 


812 


720 


1.990 


20' 


050 


631 


851 


709 


159 


980 


40' 


30' 


.5075 


.8616 


.5890 


1.698 


1.161 


1.970 


30' 


40' 


100 


601 


930 


686 


163 


961 


20' 


30° 50' 
31° 00' 

10' 


125 


587 

.8572 

557 


.5969 


675 


165 
1.167 


951 


10' 
59° 00' 

58° 50' 


.5150 


.6009 


1.664 


1.942 


175 


048 


653 


169 


932 


20' 


200 


542 


088 


643 


171 


923 


40' 


30' 


.5225 


.8526 


.6128 


1.632 


1.173 


1.914 


30' 


40' 


250 


511 


168 


621 


175 


905 


20' 


31° 50' 
32° 00' 

10' 


275 


496 


208 


611 


177 
1.179 


896 

1.887 
878 


10' 
58° 00' 

57° 50' 


.5299 


.8480 


.6249 


1.600 


324 


465 


289 


590 


181 


20' 


348 


450 


330 


580 


184 


870 


40' 


30' 


.5373 


.8434 


.6371 


1.570 


1.186 


1.861 


30' 


40' 


398 


418 


412 


560 


188 


853 


20' 


32° 50' 
33° 00' 

10' 


422 


403 


453 


550 


190 

1.192 

195 


844 


10' 
57° 00' 

56° 50' 


.5446 


.8387 


.6494 


1.540 


1.836 


471 


371 


536 


530 


828 


20' 


495 


355 


577 


520 


197 


820 


40' 


30' 


.5519 


.8339 


.6619 


1.511 


1.199 


1.812 


30' 


40' 


544 


323 


661 


501 


202 


804 


20' 


33° 50' 
34° 00' 

10' 


568 

.5592 

616 


307 


703 


492 


204 


796 


10' 
66° 00' 

55° 50' 


.8290 


.6745 


1.483 


1.206 
209 


1.788 


274 


787 


473 


781 


20' 


640 


258 


830 


464 


211 


773 


40' 


30 


.5664 


.8241 


.6873 


1.455 


1.213 


1.766 


30' 


40 


688 


225 


916 


446 


216 


758 


20' 


34° 50 
35° 00' 

10' 


712 


208 


.6959 


437 


218 


751 


10' 
55° 00' 

54° 50' 


.5736 


.8192 


.7002 


1.428 


1.221 
223 


1.743 


760 


175 


046 


419 


736 


20' 


783 


158 


089 


411 


226 


729 


40' 


30' 


.5807 


.8141 


.7133 


1.402 


1.228 


1.722 


30' 


40' 


831 


124 


177 


393 


231 


715 


20' 


35° 50' 
36° 00' 


854 


107 


221 


385 


233 


708 


10' 
54° 00' 


.5878 
Cos 


.8090 


.7265 


1.376 


1.236 
Cse 


1.701 


Sin 


Cot 


Tan 


Sec 





Appendix A 



319 







TABLE I] 


'. (continued) 








Sin 

.5878 


Cos 

.8090 


Tan 

.7265 


Cot 


See 


Csc 






36° 00' 

10' 


1.376 
368 


1.236 


1.701 


54° 00' 

53° 50' 


901 


073 


310 


230 


695 


20' 


925 


056 


355 


360 


241 


688 


40' 


30' 


.5948 


.8039 


.7400 


1.351 


1.244 


1.681 


30' 


40' 


972 


021 


445 


343 


247 


675 


20' 


36° 50' 
37° 00' 

10' 


.5995 
.6018 


.8004 


490 


335 


249 


668 


10' 
53° 00' 

52° 50' 


.7986 


.7536 


1.327 


1.252 


1.662 


041 


969 


581 


319 


255 


655 


20' 


065 


951 


627 


311 


258 


649 


4C 


30' 


.6088 


.7934 


.7673 


1.303 


1.260 


1.643 


30' 


40' 


111 


916 


720 


295 


263 


636 


20' 


37° 50' 

38° 00' 

10' 


134 


898 


766 


288 


266 


630 


10' 
52° 00' 

51° 50' 


.6157 


.7880 


.7813 


1.280 


1.269 


1.624 


180 


862 


860 


272 


272 


618 


20' 


202 


844 


907 


265 


275 


612 


40' 


30' 


.6225 


.7826 


.7954 


1.257 


1.278 


1.606 


30' 


40' 


248 


808 


.8002 


250 


281 


601 


20' 


38° 50' 

39° 00' 

10' 


271 


790 


050 


242 


284 


595 


10' 
51° 00' 

50° 50' 


.6293 


.7771 


.8098 


1.235 


1.287 


1.589 


316 


753 


146 


228 


290 


583 


20' 


338 


735 


195 


220 


293 


578 


40' 


30' 


.6361 


.7716 


.8243 


1.213 


1.296 


1.572 


30' 


40' 


383 


698 


292 


206 


299 


567 


20' 


39° 50' 

40° 00' 

10 


406 


679 


342 
.8391 


199 


302 


561 


10' 
50° 00 

49° 50' 


.6428 


.7660 


1.192 


1.305 


1.556 


450 


642 


441 


185 


309 


550 


20' 


472 


623 


491 


178 


312 


545 


40' 


30' 


.6494 


.7604 


.8541 


1.171 


1.315 


1.540 


30' 


40' 


517 


585 


591 


164 


318 


535 


20' 


40° 50' 
41° 00' 

10' 


539 


566 
.7547 

528 


642 


157 


322 


529 


10' 
49° 00' 

48° 50' 


.6561 


.8693 


1.150 


1.325 


1.524 


583 


744 


144 


328 


519 


20' 


604 


509 


796 


137 


332 


514 


40' 


30' 


.6626 


.7490 


.8847 


1.130 


1.335 


1.509 


30' 


40' 


648 


470 


899 


124 


339 


504 


20' 


41° 50' 
42° 00' 


670 


451 


.8952 


117 


342 


499 


10' 
48° 00* 


.6691 


.7431 
Sin 


.9004 
Cot 


1.111 


1.346 


1.494 




Cos 


Tan 


Csc 


Sec 







320 



Appendix A 









TABLE II 


(continued) 






i 


Sin 


Cos 


Tan 


Cot 


Sec 


Csc 






42 c 


>00' 

10' 


.6691 


.7431 


.9004 


1.111 


1.346 


1.494 


48° OC 

47° 50' 


713 


412 


057 


104 


349 


490 




20' 


734 


392 


110 


098 


353 


485 


40' 




30' 


.6756 


.7373 


.9163 


1.091 


1.356 


1.480 


30' 




40' 


777 


353 


217 


085 


360 


476 


20' 


42° 
43° 


50' 
00' 

10' 


799 


333 


271 


079 


364 


471 


47° 00' 

46° 50' 


.6820 


.7314 


.9325 


1.072 


1.367 


1.466 


841 


294 


380 


066 


371 


462 




20' 


862 


274 


435 


060 


375 


457 


40' 




30' 


.6884 


.7254 


.9490 


1.054 


1.379 


1.453 


30' 




40' 


905 


234 


545 


048 


382 


448 


20' 


43° 
44° 


50' 

w 

10' 


926 


214 


601 


042 


386 


444 


10' 
46° 00' 

45° 50' 


.6947 


.7193 


.9657 


1.036 


1.390 


1.440 


967 


173 


713 


030 


394 


435 




20' 


.6988 


153 


770 


024 


398 


431 


40' 




30* 


.7009 


.7133 


.9827 


1.018 


1.402 


1.427 


30' 




40' 


030 


112 


884 


012 


406 


423 


20' 


44° 
46° 


50' 


050 


092 
.7071 

Sin 


.9942 


006 


410 


418 


10' 
45° 00' 


.7071 
Cos 


1.000 


1.000 


1.414 
Csc 


1.414 


Cot 


Tan 


See 





Appendix A 



321 



TABLE III 
Four-Place Logarithms of Numbers 



m 

10 

n 

12 
13 

14 
15 

16 

17 
18 
19 

20 

21 
22 
23 

24 
25 

26 

27 
28 
29 

30 

31 
32 
33 

34 
35 

36 

37 
38 
39 

40 

N 





1 

0043 

0453 
0828 
1173 

1492 
1790 
2068 

2330 
2577 
2810 

3032 

3243 
3444 
3636 

3820 
3997 
4166 

4330 
4487 
4639 

4786 

4928 
5065 
5198 

5328 
5453 
5575 

5694 
5809 
5922 

6031 
1 


8 

0086 

0492 
0864 
1206 

1523 
1818 
2095 

2355 
2601 
2833 

3054 

3263 
3464 
3655 

3838 
4014 
4183 

4346 
4502 
4654 

4800 

4942 
5079 
5211 

5340 
5465 
5587 

5705 
5821 
5933 

6042 
9 


3 

0128 

0531 
0899 
1239 

1553 
1847 
2122 

2380 
2625 
2856 

3075 

3284 
3483 
3674 

3856 
4031 
4200 

4362 
4518 
4669 

4814 

4955 
5092 
5224 

5353 
5478 
5599 

5717 
5832 
5944 

6053 
3 


4 

0170 

0569 
0934 
1271 

1584 
1875 
2148 

2405 
2648 
2878 

3096 

3304 
3502 
3692 

3874 
4048 
4216 

4378 
4533 
4683 

4829 

4969 
5105 
5237 

5366 
5490 
5611 

5729 
5843 
5955 

6064 
4 


5 

0212 

0607 
0969 
1303 

1614 
1903 
2175 

2430 
2672 
2900 

3118 

3324 
3522 
3711 

3892 
4065 
4232 

4393 
4548 
4698 

4843 

4983 
5119 
5250 

5378 
5502 
5623 

5740 
5865 
5966 

6076 


6 

0263 

0646 
1004 
1335 

1644 
1931 
2201 

2455 
2695 
2923 

3139 

3345 
3541 
3729 

3909 
4082 
4249 

4409 
4564 
4713 

4857 

4997 
5132 
5263 

5391 
5514 
5635 

5752 
5866 
5977 

6086 
6 


7 

0294 

0682 
1038 
1367 

1673 
1959 
2227 

2480 
2718 
2945 

3160 

3365 
3560 
3747 

3927 
4099 
4266 

4425 
4579 
4728 

4871 

5011 
5145 
5276 

5403 
5527 
5647 

5763 
5877 
5988 

6096 
7 


8 

0334 

0719 
1072 
1399 

1703 
1987 
2253 

2504 
2742 
2967 

3181 

3386 
3579 
3766 

3945 
4116 
4281 

4440 
4594 
4742 

4886 

5024 
5159 
5289 

6416 
5539 
5658 

5775 

5888 
5999 

6107 
8 


9 

0374 

0755 
1106 
1430 

1732 
2014 
2279 

2529 
2765 
2989 

3201 

3404 
3598 
3784 

3962 
4133 
4298 

4456 
4609 
4757 

4900 

5038 
5172 
5302 

5428 
5551 
5670 

5786 
5899 
6010 

6117 
9 


.0000 


.0414 
.0792 
.1139 

.1461 
.1761 
.2041 

.2304 
.2553 
.2788 

.3010 

.3222 
.3424 
.3617 

.3802 
.3979 
.4150 

.4314 
.4472 
.4624 


.4771 


.4914 
.5051 
.5186 

.5315 
.5441 
.5563 

.5682 
.5798 
.5911 


.6021 





5 



322 



Appendix A 









TABLE III (continued) 








V 

40 

41 
42 
43 

44 
48 

46 

47 
48 
49 

80 

51 
52 
53 

54 
68 

56 

57 

58 
59 

60 

61 
62 
63 

64 
68 
66 

67 
68 
69 

70 
N 




.6021 

.6128 
.6232 
.6335 

.6435 
.6532 
.6628 

.6721 
.6812 
.6902 

.6990 


1 

6031 

6138 
6243 
6345 

6444 
6542 
6637 

6730 
6821 
6911 

6998 

7084 
7168 
7251 

7332 
7412 
7490 

7566 
7642 
7716 

7789 

7860 
7931 
8000 

8069 
8136 
8202 

8267 
8331 
8395 

8457 
1 


2 

6042 

6149 
6253 
6355 

6454 
6551 
6646 

6739 
6830 
6920 

7007 

7093 
7177 
7259 

7340 
7419 
7497 

7574 
7649 
7723 

7796 

7868 
7938 
8007 

8075 
8142 
8209 

8274 
8338 
8401 

8463 
S 


3 

6053 

6160 
6263 
6365 

6464 
6561 
6656 

6749 
6839 
6928 

7016 

7101 
7185 
7267 

7348 
7427 
7506 

7582 
7657 
7731 

7803 

7875 
7945 
8014 

8082 
8149 
8215 

8280 
8344 
8407 

8470 
8 


4 

6064 

6170 
6274 
6375 

6474 
6571 
6666 

6758 
6848 
6937 

7024 

7110 
7193 
7275 

7356 
7435 
7513 

7589 
7664 
7738 

7810 

7882 
7952 
8021 

8089 
8156 
8222 

8287 
8351 
8414 

8476 
4 


6 

6075 

6180 
3284 
6386 

6484 
6580 
6675 

6767 
6857 
6946 

7033 

7118 
7202 
7284 

7364 
7443 
7520 

7597 
7672 
7745 

7818 

7889 
7959 
8028 

8096 
8162 
8228 

8293 
8357 
8420 

8482 
6 


6 

6085 

6191 
6294 
6395 

6493 
6590 
6684 

6776 
6866 
6955 

7042 

7126 
7210 
7292 

7372 
7451 
7528 

7604 
7679 
7752 

7825 

7896 
7966 
8035 

8102 
8169 
8235 

8299 
8363 
8426 

8488 
6 


7 

6096 

6201 
6304 
6405 

6603 
6699 
6693 

6785 
6875 
6964 

7050 

7135 
7218 
7300 

7380 
7459 
7536 

7612 
7686 
7760 

7832 

7903 
7973 
8041 

8109 
8176 
8241 

8306 
8370 
8432 

8494 
7 


8 

6107 

6212 
6314 
6415 

6513 
6609 
6702 

6794 
6884 
6972 

7059 

7143 
7226 
7308 

7388 
7466 
7543 

7619 
7694 
7767 

7839 

7910 
7980 
8048 

8116 
8182 
8248 

8312 
8376 
8439 

8500 
8 


9 

6117 

6222 
6325 
6425 

6522 
6618 
6712 

6803 
6893 
6981 

7067 

7152 
7235 
7316 

7396 
7474 
7551 

7627 
7701 
7774 

7846 

7917 
7987 
8055 

8122 
8189 
8254 

8319 
8382 
8445 

8506 
9 


.7076 
.7160 
.7243 

.7324 
.7404 
.7482 

.7559 
.7634 
.7709 

.7782 


.7853 
.7924 
.7993 

.8062 
.8129 
.8195 

.8261 
.8325 
.8388 


.8451 






Appendix A 



323 



TABLE III (continued) 



N 

70 

71 
72 
73 

74 
75 

76 

77 
78 
79 

80 

81 
82 
83 

84 
85 

86 

87 
88 
89 

90 

91 
92 
93 

94 
95 

96 

97 
98 
99 

N 




.8451 


1 

8457 


S 

8463 


3 

8470 

8531 
8591 
8651 

8710 
8768 
8825 

8882' 
8938 
8993 

9047 

9101 
9154 
9206 

9258 
9309 
9360 

9410 
9460 
9509 

9557 

9605 
9652 
9699 

9745 
9791 
9836 

9881 
9926 
9969 

3 


4 
8476 


5 

8482 

8543 
8603 
8663- 

8722 
8779 
8837 

8893 
8949 
9004 

9058 

9112 
9165 
9217 

9269 
9320 
9370 

9420 
9469 
9518 


6 

8488 

8549 
8609 
8669 

8727 
8785 
8842 

8899 
8954 
9009 

9063 

9117 
9170 
9222 

9274 
9325 
9375 

9425 
9474 
9523 

9571 

9619 
9666 
9713 

9759 
9805 
9850 

9894 
9939 
9983 

6 


7 

8494 


8 

8500 

8561 
8621 
8681 

8739 
8797 
8854 

8910 
8965 
9020 

9074 

9128 
9180 
9232 

9284 
9335 
9385 

9435 
9484 
9533 

9581 


9 

8506 

8567 
8627 
8686 

8745 
8802 
8859 

8915 
8971 
9025 

9079 

9133 
9186 
9238 

9289 
9340 
9390 

9440 
9489 
9538 

9586 

9633 
9680 
9727 

9773 
9818 
9863 

9908 
9952 
9996 

9 


.8513 
.8573 
.8633 

.8692 
.8751 
.8808 

.8865 
.8921 
.8976 


8519 
8579 
8639 

8698 
8756 
8814 

8871 
8927 
8982 

9036 

9090 
9143 
9196 

9248 
9299 
9350 

9400 
9450 
9499 

9547 

9595 
9643 
9689 

9736 
9782 
9827 

9872 
9917 
9961 

1 


8525 
8585 
8645 

8704 
8762 
8820 

8876 
8932 
8987 


8537 
8597 
8657 

8716 
8774 
8831 

8887 
8943 
8998 


8555 
8615 
8675 

8733 
8791 
8848 

8904 
8960 
9015 


.9031 


9042 


9053 


9069 


.9085 
.9138 
.9191 

.9243 
.9294 
.9345 

.9395 
.9445 
.9494 


9096 
9149 
9201 

9253 
9304 
9355 

9405 
9455 
9504 


9106 
9159 
9212 

9263 
9315 
9365 

9415 
9465 
9513 


9122 
9175 
9227 

9279 
9330 
9380 

9430 
9479 
9528 

9576 

9624 
9671 
9717 

9763 
9809 
9854 

9899 
9943 
9987 

7 


.9542 


9552 

9600 
9647 
9694 

9741 
9786 
9832 

9877 
9921 
9965 

2 


9562 


9566 


.9590 
.9638 
.9685 

.9731 
.9777 
.9823 

.9868 
.9912 
.9956 


9609 
9657 
9703 

9750 
9795 
9841 

9886 
9930 
9974 

4 


9614 
9661 
9708 

9754 
9800 
9845 

9890 
9934 
9978 

5 


9628 
9675 
9722 

9768 
9814 
9859 

9903 
9948 
9991 

8 






324 



Appendix A 



TABLE IV 
Four-Place Logarithms of Functions 



w 


L Sin 


L Tan 


LCot 


L Cos 

10.0000 




0°00' 


90° 00' 


10' 


7.4637 


7.4637 


12.5363 


.0000 


89° 50' 


20' 


.7648 


.7648 


.2352 


.0000 


40' 


30' 


7.9408 


7.9409 


12.0591 


.0000 


30' 


40' 


8.0658 


8.0658 


11.9342 


.0000 


20' 


0°50' 


.1627 


.1627 


.8373 


10.0000 


10' 


1°00' 


8.2419 


8.2419 


11.7581 


9.9999 


89° 00' 


10' 


.3088 


.3089 


.6911 


.9999 


88° 50' 


20' 


.3668 


.3669 


.6331 


.9999 


40' 


30' 


.4179 


.4181 


.5819 


.9999 


30' 


40' 


.4637 


.4638 


.5362 


.9998 


20' 


1° 50' 


.5050 


.5053 


.4947 


.9998 


10' 


a oo' 


8.5428 


8.5431 


11.4569 


9.9997 


88° 00' 


10' 


.5776 


.5779 


.4221 


.9997 


87° 50' 


20' 


.6097 


.6101 


.3899 


.9996 


40' 


30' 


.6397 


.6401 


.3599 


.9996 


30' 


40' 


.6677 


.6682 


.3318 


.9995 


20' 


2° 50' 


.6940 


.6945 


.3055 


.9995 


10' 


3° 00' 


8.7188 


8.7194 


11.2806 


9.9994 


87° 00' 


10' 


.7423 


.7429 


.2571 


.9993 


86° 50' 


20' 


.7645 


.7652 


.2348 


.9993 


40' 


30' 


.7857 


.7865 


.2135 


.9992 


30' 


40' 


.8059 


.8067 


.1933 


.9991 


20' 


3° 50' 


.8251 


.8261 


.1739 


.9990 


10' 


4° 00' 


8.8436 


8.8446 


11.1554 


9.9989 


86° 00' 


10' 


.8613 


.8624 


.1376 


.9989 


85° 50' 


20' 


.8783 


.8795 


.1205 


.9988 


40' 


30' 


.8946 


.8960 


.1040 


.9987 


30' 


40' 


.9104 


.9118 


.0882 


.9986 


20' 


4° 50' 


.9256 


.9272 


.0728 


.9985 


10' 


5° 00' 


8.9403 


8.9420 


11.0580 


9.9983 


85° 00' 


10' 


.9545 


.9563 


.0437 


.9982 


84° 50' 


20' 


.9682 


.9701 


.0299 


.9981 


40' 


30' 


.9816 


.9836 


.0164 


.9980 


30' 


40' 


8.9945 


8.9966 


11.0034 


.9979 


20' 


5° 50' 


9.0070 


9.0093 


10.9907 


.9977 


10' 


6° 00' 


9.0192 


9.0216 


10.9784 


9.9976 


84° 00' 




L Cos 


L Cot 


L Tan 


LSin 


„^ „„ _,. 





Appendix A 



325 



TABLE IV (continued) 



. ^ 


L Sin 

9.0192 


L Tan 

9.0216 
.0336 


L Cot 

10.9784 
.9664 


L Cos 




6° 00' 

10' 


9.9976 
.9975 


84° 00' 

83° 50' 


.0311 


20' 


.0426 


.0453 


.9547 


.9973 


40' 


30' 


.0539 


.0567 


.9433 


.9972 


30' 


40' 


.0648 


.0678 


.9322 


.9971 


20' 


6° 50' 


.0755 


.0786 


.9214 


.9969 


10' 


7° 00' 


9.0859 


9.0891 


10.9109 


9.9968 


83° 00' 


10' 


.0961 


.0995 


.9005 


.9966 


82° 50' 


20' 


.1060 


.1096 


.8904 


.9964 


40' 


30' 


.1157 


.1194 


.8806 


.9963 


30' 


40' 


.1252 


.1291 


.8709 


.9961 


20' 


7° 50' 


.1345 


.1385 


.8615 


.9959 


10' 


8° 00' 


9.1436 


9.1478 


10.8522 


9.9958 


82° 00' 


10' 


.1525 


.1569 


.8431 


.9956 


81° 50' 


20' 


.1612 


.1658 


.8342 


.9954 


40' 


30' 


.1697 


.1745 


.8255 


.9952 


30' 


40' 


.1781 


.1831 


.8169 


.9950 


20' 


8° 50' 


.1863 


.1915 


.8085 


.9948 


10' 


9° 00' 


9.1943 


9.1997 


10.8003 


9.9946 


81° 00' 


10' 


.2022 


.2078 


.7922 


.9944 


80° 50' 


20' 


.2100 


.2158 


.7842 


.9942 


40' 


30' 


.2176 


.2236 


.7764 


.9940 


30' 


40' 


.2251 


.2313 


.7687 


.9938 


20' 


9° 50' 


.2324 


.2^89 


.7611 


.9936 


10' 


10° 00' 


9.2397 


9.2463 


10.7537 


9.9934 


80° 00' 


10' 


.2468 


.2536 


.7464 


.9931 


79° 50' 


20' 


.2538 


.2609 


.7391 


.9929 


40' 


30' 


.2606 


.2680 


.7320 


.9927 


30' 


40' 


.2674 


.2750 


.7250 


.9924 


20' 


10° 50' 


.2740 


.2819 


.7181 


.9922 


10' 


11° 00' 


9.2806 


9.2887 


10.7113 


9.9919 


79° 00' 


10' 


.2870 


.2953 


.7047 


.9917 


78° 50' 


20' 


.2934 


.3020 


.6980 


.9914 


40' 


30' 


.2997 


.3085 


.6915 


.9912 


30' 


40' 


.3058 


.3149 


.6851 


.9909 


20' 


11° 50' 


.3119 


.3212 


.6788 


.9907 


10'. 


n°oo' 


9.3179 


9.3275 


10.6725 


9.9904 


78° 00' 




L Cos 


L Cot 


L Tan 


L Sin 


4 





326 



Appendix A 



TABLE IV (continued) 



V 


LSia 

9.3179 


L Tan 

9.3275 


LCot 

10.6725 


LCos 

9.9904 




12° 00' 


78° 00' 


10' 


.3238 


.3336 


.6664 


.9901 


77° 50' 


20' 


.3296 


.3397 


.6603 


.9899 


40' 


30' 


.3353 


.3458 


.6542 


.9896 


30' 


40' 


.3410 


.3517 


.6483 


.9893 


20' 


12° 50' 


.3466 


.3576 


.6424 


.9890 


10' 


13° 00' 


9.3521 


9.3634 


10.6366 


9.9887 


77° 00' 


10' 


.3575 


.3691 


.6309 


.9884 


76° 50' 


20' 


.3629 


.3748 


.6252 


.9881 


40' 


30' 


.3682 


.3804 


.6196 


.9878 


30' 


40' 


.3734 


.3859 


.6141 


.9875 


20' 


13° 50' 


.3786 


.3914 


.6086 


.9872 


10' 


14° 00' 


9.3837 


9.3968 


10.6032 


9.9869 


76° 00' 


10' 


.3887 


.4021 


.5979 


.9866 


75° 50' 


20' 


.3937 


.4074 


.5926 


.9863 


40' 


30' 


.3986 


.4127 


.5873 


.9859 


30' 


40' 


.4035 


.4178 


.5822 


.9856 


20' 


14° 50' 


.4083 


.4230 


.5770 


.9853 


10' 


15° 00' 


9.4130 


9.4281 


10.5719 


9.9849 


75° 00' 


10' 


.4177 


.4331 


.5669 


.9846 


74° 50' 


20' 


.4223 


.4381 


.5619 


.9843 


40' 


30' 


.4269 


.4430 


.5570 


.9839 


30' 


40' 


.4314 


.4479 


.5521 


.9836 


20' 


15° 50' 


.4359 


.4527 


.5473 


.9832 


10' 


16° 00' 


9.4403 


9.4575 


10.5425 


9.9828 


74° 00' 


10' 


.4447 


.4622 


.5378 


.9825 


73° 50' 


20' 


.4491 


.4669 


.5331 


.9821 


40' 


30' 


.4533 


.4716 


.5284 


.9817 


30' 


40' 


.4576 


.4762 


.5238 


.9814 


20' 


16° 50' 


.4618 


.4808 


.5192 


.9810 


10' 


17° 00' 


9.4659 


9.4853 


10.5147 


9.9806 


73° 00' 


10' 


.4700 


.4898 


.5102 


.9802 


72° 50' 


20' 


.4741 


.4943 


.5057 


.9798 


40' 


30' 


.4781 


.4987 


.5013 


.9794 


30' 


40' 


.4821 


.5031 


.4969 


.9790 


20' 


17° 50' 


.4861 


.5075 


.4925 


.9786 


10' 


18° 00' 


9.4900 
L Cos 


9.5118 
LCot 


10.4882 
LTan 


9.9782 
LSin 


72° 00' 







Appendix A 



327 



TABLE IV (continued) 



, i... V. 


LSin 


I. Tan 

9.5118 


LCot 

10.4882 


L Cos 

9.9782 




I... pm 

18° 00' 


9.4900 


72° 00' 


10' 


.4939 


.5161 


.4839 


.9778 


71° 50' 


20' 


.4977 


.5203 


.4797 


.9774 


40' 


30' 


.5015 


.5245 


.4755 


.9770 


30' 


40' 


.5052 


.5287 


.4713 


.9765 


20' 


18° 50' 


.5090 


.5329 


.4671 


.9761 


10' 


19° 00' 


9.5126 


9.5370 


10.4630 


9.9757 


71° 00' 


10' 


.5163 


.5411 


.4589 


.9752 


70° 50' 


20' 


.5199 


.5451 


.4549 


.9748 


40' 


30' 


.5235 


.5491 


.4509 


.9743 


30' 


40' 


.5270 


.5531 


.4469 


.9739 


20' 


19° 50' 


.5306 


.5571 


.4429 


.9734 


10' 


20° 00' 


9.5341 


9.5611 


10.4389 


9.9730 


70° 00' 


10' 


.5375 


.5650 


.4350 


.9725 


69° 50' 


20' 


.5409 


.5689 


.4311 


.9721 


40' 


30' 


.5443 


.5727 


.4273 


.9716 


• 30' 


40' 


.5477 


.5766 


.4234 


.9711 


20' 


20° 50' 


.5510 


.5804 


.4196 


.9706 


10* 


21° 00' 


9.5543 


9.5842 


10.4158 


9.9702 


69° OC 


10' 


.5576 


.5879 


.4121 


.9697 


68° sty 


20' 


.5609 


.5917 


.4083 


.9692 


Aff 


30' 


.5641 


.5954 


.4046 


.9687 


30* 


40' 


.5673 


.5991 


.4009 


.9682 


20' 


21° 50' 


.5704 


.6028 


.3972 


.9677 


10' 


22° 00' 


9.5736 


9.6064 


10.3936 


9.9672 


68° 00' 


10' 


.5767 


.6100 


.3900 


.9667 


67° 50' 


20' 


.5798 


.6136 


.3864 


.9661 


40' 


30' 


.5828 


.6172 


.3828 


.9656 


30' 


40' 


.5S59 


.6208 


.3792 


.9651 


20' 


22° 50' 


.5889 


.6243 


.3757 


.9646 


10' 


23° 00' 


9.5919 


9.6279 


10.3721 


9.9640 


67° 00' 


10' 


.5948 


.6314 


.3686 


.9635 


66° 50' 


20' 


.5978 


.6348 


.3652 


.9629 


40' 


30' 


.6007 


.6383 


.3617 


.9624 


30' 


40' 


.6036 


.6417 


.3583 


.9618 


20' 


23° 50' 


.6065 


.6452 


.3548 


.9613 


10' 


24° 00' 


9.6093 
L Cos 


9.6486 
LCot 


10.3514 
LT*n 


9.9607 
LSin 


66° 00' 







328 



Appendix A 



TABLE IV (continued) 





L Sin ( 


L Tan 


LCot 
10.3514 


L Cos 


66*00' 




24*00' 


9.0093 


9.6486 


9.9607 


10' 


.6121 


.6520 


.3480 


.9602 


65* 50' 


20' 


.6149 


.6553 


.3447 


.9596 


40' 


30' 


.6177 


.6587 


.3413 


.9590 


30' 


40' 


.6205 


.6620 


.3380 


.9584 


20' 


24° 50' 


.6232 


.6654 


.3346 


.9579 


10' 


25° 00' 


9.6259 


9.6687 


10.3313 


9.9573 


66*00' 


10' 


.6286 


.6720 


.3280 


.9567 


64*50' 


20' 


.6313 


.6752 


.3248 


.9561 


40' 


30' 


.6340 


.6785 


.3215 


.9555 


30' 


40' 


.6366 


.6817 


.3183 


.9549 


20' 


25° 50' 


.6392 


.6850 


.3150 


.9543 


10' 


2$° 00' 


9.6418 


9.6882 


10.3118 


9.9537 


64° 00' 


10' 


.6444 


.6914 


.3086 


.9530 


63* 50' 


20' 


.6470 


.6946 


.3054 


.9524 


40' 


30' 


.6495 


.6977 


.3023 


.9518 


30' 


40' 


.6521 


.7009 


.2991 


.9512 


20' 


26* 50' 


.6546 


.7040 


.2960 


.9505 


10' 


27*00' 
10' 


9.6570 


9.7072 
.7103 


10.2928 
.2897 


9.9499 
.9492 


63*00' 

62* 50' 


.6595 


20' 


.6620 


.7134 


.2866 


.9486 


40' 


30' 


.6644 


.7165 


.2835 


.9479 


30' 


40' 


.6668 


.7196 


.2804 


.9473 


20' 


27* 50' 


.6692 


.7226 


.2774 


.9466 


10' 


28*00' 


9.6716 


9.7257 


10.2743 


9.9459 


62*00' 


10' 


.6740 


.7287 


.2713 


.9453 


61* 50' 


20' 


.6763 


.7317 


.2683 


.9446 


40' 


30' 


.6787 


.7348 


.2652 


.9439 


30' 


40' 


.6810 


.7378 


.2622 


.9432 


20' 


28* 50' 
29*00' 

10' 


.6833 


.7408 

9.7438 

.7467 


.2592 

10.2562 

.2533 


.9425 

9.9418 

.9411 


10' 
61*00' 

60*50' 


9.6856 


.6878 


20' 


.6901 


.7497 


.2503 


.9404 


40' 


30' 


.6923 


.7526 


.2474 


.9397 


30* 


40' 


.6946 


.7556 


.2444 


.9390 


2C 


29*50' 
30*00' 


.6968 


.7585 
9.7614 


.2415 

10.2386 

LTan 


.9383 
9.9375 
L Sin 


lC 
60*00' 


9.6990 




LCos 


LCot 





Appendix A 
TABLE IV {continued) 



329 





L Sin 


L Tan 


LCot 


L Cos 






30° 00' 


9.6990 


9.7614 


10.2386 


9.9375 


W ° 00' 


10' 


.7012 


.7644 


.2356 


.9368 


59° 50' 


20' 


.7033 


.7673 


.2327 


.9361 


40' 


30' 


.7055 


.7701 


.2299 


.9353 


30' 


40' 


.7076 


.7730 


.2270 


.9346 


20' 


30° 50' 


.7097 


.7759 


.2241 


.9338 


10' 


31° 00' 


9.7118 


9.7788 


10.2212 


9.9331 


59° 00' 


10' 


.7139 


.7816 


.2184 


.9323 


58° 50' 


20' 


.7160 


.7845 


.2155 


.9315 


40' 


30' 


.7181 


.7873 


.2127 


.9308 


30' 


40' 


.7201 


.7902 


.2098 


.9300 


20' 


31° 50' 


.7222 


.7930 


.2070 


.9292 


10' 


32° 00' 


9.7242 


9.7958 


10.2042 


9.9284 


58° 00' 


10' 


.7262 


.7986 


.2014 


.9276 


57° 50' 


20' 


.7282 


.8014 


.1986 


.9268 


40' 


30' 


.7302 


.8042 


.1958 


.9260 


30' 


40' 


.7322 


.8070 


.1930 


.9252 


20' 


32° 50' 


.7342 


.8097 


.1903 


.9244 


10' 


33° 00' 


9.7361 


9.8125 


10.1875 


9.9236 


57° 00' 


10' 


.7380 


.8153 


.1847 


.9228 


56° 50' 


20' 


.7400 


.8180 


.1820 


.9219 


40' 


30' 


.7419 


.8208 


.1792 


.9211 


30' 


40' 


.7438 


.8235 


.1765 


.9203 


20' 


33° 50' 


.7457 


.8263 


.1737 


.9194 


10' 


34° 00' 


9.7476 


9.8290 


10.1710 


9.9186 


56° 0C 


10' 


.7494 


.8317 


.1683 


.9177 


55° 50' 


20' 


.7513 


.8344 


.1656 


.9169 


40' 


30' 


.7531 


.8371 


.1629 


.9160 


30' 


40' 


.7550 


.8398 


.1602 


.9151 


20' 


34° 50' 


.7568 


.8425 


.1575 


.9142 


10' 


35° 00' 

10' 


9.7586 


9.8452 
.8479 


10.1548 
.1521 


9.9134 
.9125 


55° 00' 

54° 50' 


.7604 


20' 


.7622 


.8506 


.1494 


.9116 


40' 


30' 


.7640 


.8533 


.1467 


.9107 


30' 


40' 


.7657 


.8559 


.1441 


.9098 


20' 


35° 50' 


.7675 


-8586 


.1414 


.9089 


10' 


36° 00' 


9.7692 
LCos 


9.8613 
LCot 


10.1387 
LTan 


9.9080 


64° 00' 




LSin 







330 



Appendix A 



TABLE IV (continued) 



, ^ 


LSin 


LTan 


LCot 


L Cos 






36° 00' 

10' 


9.7692 
.7710 


9.8613 
.8639 


10.1387 
.1361 


9.9080 


54° 00' 

53° 50' 


.9070 


20' 


.7727 


.8666 


.1334 


.9061 


40' 


30' 


.7744 


.8692 


.1308 


.9052 


30' 


40' 


.7761 


.8718 


.1282 


.9042 


20' 


36° 50' 
37° 00' 


.7778 
9.7795 


.8745 
9.8771 


.1255 
10.1229 


.9033 


10' 
53° 00' 


9.9023 


10' 


.7811 


.8797 


.1203 


.9014 


52° 50' 


20' 


.7828 


.8824 


.1176 


.9004 


40' 


30' 


.7844 


.8850 


.1150 


.8995 


30' 


40' 


.7861 


.8876 


.1124 


.8985 


20' 


37° 50' 


.7877 


.8902 


.1098 


.8975 


10' 


38° 00' 


9.7893 


9.8928 


10.1072 


9.8965 


52° 00' 


10' 


.7910 


.8954 


.1046 


.8955 


51° 50' 


20' 


.7926 


.8980 


.1020 


.8945 


40' 


30' 


.7941 


.9006 


.0994 


.8935 


30' 


40' 


.7957 


.9032 


.0968 


.8925 


20' 


38° 50' 


.7973 


.9058 


.0942 


.8915 


10' 


39° 00' 


9.7989 


9.9084 


10.0916 


9.8905 


51° 00' 


10' 


.8004 


.9110 


.0890 


.8895 


50° 50' 


20' 


.8020 


.9135 


.0865 


.8884 


40' 


30' 


.8035 


.9161 


.0839 


.8874 


30' 


40' 


.8050 


.9187 


.0813 


.8864 


20' 


39° 50' 


.8066 


.9212 


.0788 


.8853 


10' 


40° 00' 


9.8081 


9.9238 


10.0762 


9.8843 


50° 00' 


10' 


.8096 


.9264 


.0736 


.8832 


49° 50' 


20' 


.8111 


.9289 


.0711 


.8821 


40' 


30' 


.8125 


.9315 


.0685 


.8810 


30' 


40' 


.8140 


.9341 


.0659 


.8800 


20' 


40° 50' 


.8155 


.9366 


.0634 


.8789 


10' 


41° 00' 


9.8169 


9.9392 


10.0608 


9.8778 


49° 00' 


10' 


.8184 


.9417 


.0583 


.8767 


48° 50' 


20' 


.8198 


.9443 


.0557 


.8756 


40' 


30' 


.8213 


.9468 


.0532 


.8745 


30' 


40' 


.8227 


.9494 


.0506 


.8733 


20' 


41° 50' 


.8241 


.9519 


.0481 


.8722 


10' 


42° 00' 


9.8255 
LCos 


9.9544 


10.0456 


9.8711 


48° 00' 




LCot 


L tan 


LSin 


^ 





Appendix A 



331 



TABLE IV (continued) 



^ 


LSin 


LTan 


L Cot 


L Cos 

9.8711 






42° 00' 


9.8255 


9.9544 


10.0456 


48° 00' 


10' 


.8269 


.9570 


.0430 


.8699 


47° 50' 


20' 


.8283 


.9595 


.0405 


.8688 


40' 


30' 


.8297 


.9621 


.0379 


.8676 


30' 


40' 


.8311 


.9646 


.0354 


.8665 


20' 


42° 50' 


.8324 


.9671 


.0329 


.8653 


10' 


43° 00' 


9.8338 


9.9697 


10.0303 


9.8641 


47° 00' 


10' 


.8351 


.9722 


.0278 


.8629 


46° 50' 


20' 


.8365 


.9747 


.0253 


.8618 


40' 


30' 


.8378 


.9772 


.0228 


.8606 


30' 


40' 


.8391 


.9798 


.0202 


.8594 


20' 


43° 50' 


.8405 


.9823 


.0177 


.8582 


10' 


44° 00' 


9.8418 


9.9848 


10.0152 


9.8569 


46° 00' 


10' 


.8431 


.9874 


.0126 


.8557 


45° 50' 


20' 


.8444 


.9899 


.0101 


.8545 


40' 


30' 


.8457 


.9924 


.0076 


.8532 


30' 


40' 


.8469 


.9949 


.0051 


.8520 


20' 


44° 50' 
45° 00' 


.8482 
9.8495 

L Cos 


.9975 


.0025 
10.0000 


.8507 
9.8495 


10' 
45° 00' 


10.0000 




LCot 


LTan 


LSin 





332 



Appendix A 



TABLE V 
Squares and Square Roots 



N 


N* 


Vn 


VlON 


N 


AP 


Vn 


VlON 


N 


N* 


Vn 


VlON 


1.00 


1.0000 


1.00000 


3.16228 


1.60 


2.5600 


1.26491 


4.00000 


2.80 


4.8400 


1.48324 


4.69042 


1.01 


1.0201 


1.00499 


3.17805 


1.61 


2.5921 


1.26886 


4.01248 


2.21 


4.8841 


1.48661 


4.70106 


1.02 


1.0404 


1.00995 


3.19374 


1.62 


2.6244 


1.27279 


4.02492 


2.22 


4.9284 


1.48997 


4.71169 


1.03 


1.0609 


1.01489 


3.20936 


1.63 


2.6569 


1.27671 


4.03733 


2.23 


4.9729 


1.49332 


4.72229 


1.04 


1.0816 


1.01980 


3.22490 


1.64 


2.6896 


1.28062 


4.04969 


2.24 


5.0176 


1.49666 


4.73286 


1.05 


1.1025 


1.02470 


3.24037 


1.65 


2.7225 


1.28452 


4.06202 


2.25 


5.0625 


1.50000 


4.74342 


1.06 


1.1236 


1.02956 


3.25576 


1.66 


2.7556 


1.28841 


4.07431 


2.26 


5.1076 


1.50333 


4.75395 


1.07 


1.1449 


1.03441 


3.27109 


1.67 


2.7889 


1.29228 


4.08656 


2.27 


5.1529 


1.50665 


4.76445 


1.08 


1.1664 


1.03923 


3.28634 


1.68 


2.8224 


1.29615 


4.09878 


2.28 


5.1984 


1.50997 


4.77493 


1.09 


1.1881 


1.04403 


3.30151 


1.69 


2.8561 


1.30000 


4.11096 


2.29 


5.2441 


1.51327 


4.78539 


1.10 

1.11 


1.2100 


1.04881 


3.31662 


1.70 

1.71 


2.8900 


1.30384 


4.12311 


2.30 

2.31 


'5.2900 


1.51658 


4.79583 


1.2321 


1.05357 


3.33167 


2.9241 


1.30767 


4.13521 


5.3361 


1.51987 


4.80625 


1.12 


1.2544 


1.05830 


3.34664 


1.72 


2.9584 


1.31149 


4.14729 


2.32 


6.3824 


1.52316 


4.81664 


1.13 


1.2769 


1.06301 


3.36155 


1.73 


2.9929 


1.31529 


4.15933 


2.33 


5.4289 


1.52643 


4.82701 


1.14 


1.2996 


1.06771 


3.37639 


1.74 


3.0276 


1.31909 


4.17133 


2.34 


5.4756 


1.52971 


4.83735 


1.15 


1.3225 


1.07238 


3.39116 


1.75 


3.0625 


1.32288 


4.18330 


2.35 


5.5225 


1.53297 


4.84768 


1.16 


1.3456 


1.07703 


3.40588 


1.76 


3.0976 


1.32665 


4.19524 


2.36 


5.5696 


1.53623 


4.85798 


1.17 


1.3689 


1.08167 


3.42053 


1.77 


3.1329 


1.33041 


4.20714 


2.37 


5 6169 


1.53948 


4.86826 


1.18 


1.3924 


1.08628 


3.43511 


1.78 


3.1684 


1.33417 


4.21900 


2.38 


5.6644 


1.54272 


4.87852 


1.19 


1.4161 


1.09087 


3.44964 


1.79 


3.2041 


1.33791 


4.23084 


2.39 


5.7121 


1.54596 


4.88876 


1J0 

1.21 


1.4400 


1.09545 


3.46410 


1.80 

1.81 


3.2400 


1.34164 


4.24264 


2.40 

2.41 


5.7600 


1.54919 


4.89898 


1.4641 


1.10000 


3.47851 


3.2761 


1.34536 


4.25441 


5.8081 


1.55242 


4.90918 


1.22 


1.4884 


1.10454 


3.49285 


1.82 


3.3124 


1.34907 


4.26615 


2.42 


5.8564 


1.65663 


4.91935 


1.23 


1.5129 


1.10905 


3.50714 


1.83 


3.3489 


1.35277 


4.27785 


2.43 


5.9049 


1.55885 


4.92950 


1.24 


1.5376 


1.11355 


3.52136 


1.84 


3.3856 


1.35647 


4.28952 


2.44 


5.9536 


1.56205 


4.93964 


1.15 


1.5625 


1.11803 


3.53553 


1.85 


3.4225 


1.36015 


4.30116 


2.45 


6.0025 


1.56525 


4.94976 


1.26 


1.6876 


1.12250 


3.54965 


1.86 


3.4596 


1.36382 


4.31277 


2.46 


6.0516 


1.56844 


4.95984 


1.27 


1.6129 


1.12694 


3.56371 


1.87 


3.4969 


1.36748 


4.32435 


2.47 


6.1009 


1.57162 


4.96991 


1.28 


1.6384 


1.13137 


3.57771 


1.88 


3.5344 


1.37113 


4.33590 


2.48 


6.1504 


1.57480 


4.97996 


1.29 


1.6641 


1.13578 


3.59166 


1.89 


3.6721 


1.37477 


4.34741 


2.49 


6.2001 


1.57797 


4.98999 


1.80 

1.31 


1.6900 


1.14018 


3.60555 


1.90 
1.91 


3.6100 


1.37840 


4.35890 


2.50 

2.51 


6.2500 


1.68114 


6.00000 


1.7161 


1.14455 


3.61939 


3.6481 


1.38203 


4.37035 


6.3001 


1.58430 


5.00999 


1.32 


1.7424 


1.14891 


3.63318 


1.92 


3.6864 


1.38564 


4.38178 


2.52 


6.3504 


1.58745 


6.01996 


1.33 


1.7689 


1.15326 


3.64692 


1.93 


3 7249 


1.38924 


4.39318 


2.53 


6.4009 


1.59060 


6.02991 


1.34 


1.7950 


1.15758 


3.66060 


1.94 


3.7036 


1.39284 


4.40454 


2.54 


6.4516 


1.59374 


5.03984 


1.35 


1.8225 


1.16190 


3.67423 


1.95 


3.8025 


1.39642 


4.41588 


2.55 


6.5025 


1.59687 


5.04975 


1.36 


1.8496 


1.16619 


3.68782 


1.96 


3.8416 


1.40000 


4.42719 


2.56 


6.5536 


1.60000 


5.05964 


1.37 


1.8769 


1.17047 


3.70135 


1.97 


3.8809 


1.40357 


4.43847 


2.57 


6.6049 


1.60312 


5.06952 


1.38 


1.9044 


1.17473 


3.71484 


1.98 


3.9204 


1.40712 


4.44972 


2.58 


6.6564 


1.60624 


5.07937 


1.39 


1.9321 


1.17898 


3.72827 


1.99 


3.9601 


1.41067 


4.46094 


2.59 


6.7081 


1.60935 


5.08920 


1.40 

1.41 


1.9600 


1.18322 


3.74166 


2.00 

2.01 


4.0000 


1.41421 


4.47214 


2.60 

2.61 


6.7600 


1.61245 


5.09902 


1.9881 


1.18743 


3.75500 


4.0401 


1.41774 


4.48330 


6.8121 


1.61555 


5.10882 


1.42 


2.0164 


1.19164 


3.76829 


2.02 


4.0804 


1.42127 


4.49444 


2.62 


6.8644 


1.61864 


6.11859 


1.43 


2.0449 


1.19583 


3.78153 


2.03 


4.1209 


1.42478 


4.50555 


2.63 


6.9169 


1.62173 


6.12835 


1.44 


2.0736 


1.20000 


3.79473 


2.04 


4.1616 


1.42829 


4.51664 


2.64 


6.9696 


1.62481 


5.13809 


1.45 


2.1025 


1.20416 


3.80789 


8.06 


4.2025 


1.43178 


4.52769 


2.65 


7.0225 


1.62788 


6.14782 


1.46 


2.1316 


1.20830 


3.82099 


2.06 


4.2436 


1.43527 


4.53872 


2.66 


7.0756 


1.63095 


5.15752 


1.47 


2.1609 


1.21244 


3.83406 


2.07 


4.2849 


1.43875 


4.54973 


2.67 


7.1289 


1.63401 


5.16720 


1.48 


2.1904 


1.21655 


3.84708 


2.08 


4.3264 


1.44222 


4.56070 


2.68 


7.1824 


1.63707 


5.17687 


1.49 


2.2201 


1.22066 


3.86005 


2.09 


4.3681 


1.44568 


4.57165 


2.69 


7.2361 


1.64012 


5.18652 


1.60 

1.51 


2.2500 


1.22474 


3.87298 


1.10 

2.11 


4.4100 


1.44914 


4.58258 


2.70 

2.71 


7.2900 


1.64317 


6.19615 


2.2801 


1.22882 


3.88587 


4.4521 


1.45258 


4.59347 


7.3441 


1.64621 


5.20577 


1.52 


2.3104 


1.23288 


3.89872 


2.12 


4.4944 


1.45602 


4.60435 


2.72 


7.3984 


1.64924 


5.21636 


1.53 


2.3409 


1.23693 


3.91152 


2.13 


4.5369 


1.45945 


4.61519 


2.73 


7.4529 


1.66227 


5.22494 


1.54 


2.3710 


1.24097 


3.92428 


2.14 


4.5796 


1.46287 


4.62601 


2.74 


7.5076 


1.65529 


5.23450 


1.55 


2.4025 


1.24499 


3.93700 


8.15 


4.6225 


1.46629 


4.63681 


2.75 


7.5625 


1.65831 


5.24404 


1.56 


2.4336 


1.24900 


3.94968 


2.16 


4.6656 


1.46969 


4.64758 


2.76 


7.6176 


1.66132 


5.25357 


1.57 


2.4649 


1.25300 


3.96232 


2.17 


4.7089 


1.47309 


4.65833 


2.77 


7.6729 


1.66433 


5.26308 


1.58 


2.4964 


1.25698 


3.97492 


2.18 


4.7524 


1.47648 


4.66905 


2.78 


7.7284 


1.66733 


5.27257 


1.59 


2.5281 


1.26095 


3.98748 


2.19 


4.7961 


1.47986 


4.67974 


2.79 


7.7841 


1.67033 


5.28205 


1.50 
N 


2.5600 


1.26491 


4.00000 


2.20 


4.8400 


1.48324 


4.69042 


2.80 


7.8400 


1.67332 


5.29150 


N* 


Vn 


Vvon 


N 


N* 


Vn 


VlON 


N 


- N» 


Vn 


VlON 



Appendix A 



333 











TA 


BLE V (cont 


inued) 










N 


N» 


Vn 


VlON 


N 
8.40 


N* 


Vn 


VlON 


1 

N 
4.00 


N* 


Vn 


Vmr 


1.80 


7.8400 


1.67332 


5.29150 


11.5600 


1.84391 


5.83095 


16.0000 


2.00000 


6.32456 


2.81 


7.8961 


1.67631 


5.30094 


3.41 


11.6281 


1.84662 


5.83952 


4.01 


16.0801 


2.00250 


6.33246 


2.82 


7.9524 


1.67929 


5.31037 


3.42 


11.6964 


1.84932 


5.84808 


4.02 


16.1604 


2.00499 


6.34035 


2.83 


8.0089 


1.68226 


5.31977 


3.43 


11.7649 


1.85203 


5.85662 


4.03 


16.2409 


2.00749 


6.34823 


2.84 


8.0656 


1.68523 


5.32917 


3.44 


11.8336 


1.85472 


5.86515 


4.04 


16.3216 


2.00998 


6.35610 


1.65 


8.1225 


1.68819 


5.33854 


3.48 


11.9025 


1.85742 


5.87367 


44)6 


16.4025 


2.01246 


6.36396 


2.86 


8.1796 


1.69115 


5.34790 


3.46 


11.9716 


1.86011 


5.88218 


4.06 


16.4836 


2.01494 


6.37181 


2.87 


8.2369 


1.69411 


5.35724 


3.47 


12.0409 


1.86279 


5.89067 


4.07 


16.5649 


2.01742 


6.37966 


2.88 


8.2944 


1.69706 


5.36656 


3.48 


12.1104 


1.86548 


5.89915 


4.08 


16.6464 


2.01990 


6.38749 


2.89 


8.3521 


1.70000 


5.37587 


3.49 


12.1801 


1.86815 


5.90762 


4.09 


16.7281 


2.02237 


6.39531 


1.90 
2.91 


8.4100 


170294 


5.38516 


3.60 

3.51 


12.2500 


1.87083 


5.91608 


4.10 

4.11 


16.8100 


2.02485 


6.40312 


8.4681 


1.70587 


5.39444 


12.3201 


1.87350 


5.92453 
6.9&06 


16.8921 


2.02731 


6.41093 


2.92 


8.5264 


1.70880 


5.40370 


3.52 


12.3904 


1.87617 


4.12 


16.9744 


2.02978 


6.41872 


2.93 


8.5849 


1.71172 


5.41295 


3.53 


12.4609 


1.87883 


5.94138 


4.13 


17.0569 


2.03224 


6.42651 


2.94 


8.6436 


1.71464 


5.42218 


3.54 


12.5316 


1.88149 


5.94979 


4.14 


17.1396 


2.03470 


6.43428 


1.95 


8.7025 


1.71756 


5.43139 


3.55 


12.6025 


1.88414 


5.95819 


4.15 


17.2225 


2.03715 


6.44205 


2.96 


8.7616 


1.72047 


5.44059 


3.56 


12.6736 


1.88680 


5.96657 


4.16 


17.3056 


2.03961 


6.44981 


2.97 


8.8209 


1.72337 


5.44977 


3.57 


12.7449 


1.88944 


5.97495 


4.17 


17.3889 


2.04206 


6.45755 


2.98 


8.8804 


1.72627 


5.45894 


3.58 


12.8164 


1.89209 


5.98331 


4.18 


17.4724 


2.04450 


6.46529 


2.99 


8.9401 


1.72916 


5.46809 


3.59 


12.8881 


1.89473 


5.99166 


4.19 


17.5561 


2.04695 


6.47302 


3.00 

3.01 


9.0000 


1.73205 


5.47723 


3.60 

3.61 


12.9600 


1.89737 


6.00000 


4J0 

4.21 


17.6400 


2.04939 


6.48074 


9.0601 


1.73494 


5.48635 


13.0321 


1.90000 


6.00833 


17.7241 


2.05183 


6.48845 


3.02 


9.1204 


1.73781 


5.49545 


3.62 


13.1044 


1.90263 


0.01664 


4.22 


17.8084 


2.05426 


6.49615 


3.03 


9.1809 


1.74069 


5.50454 


3.63 


13.1769 


1.90526 


6.02496 


4.23 


17.8929 


2.05670 


6.50384 


3.04 


9.2416 


1.74356 


5.51362 


3.64 


13.2496 


1.90788 


6.03324 


4.24 


17.9776 


2.05913 


6.51153 


3.05 


9.3025 


1.74642 


5.52268 


3.66 


13.3225 


1.91050 


6.04152 


4J5 


18.0625 


2.06155 


6.51920 


3.06 


9.3636 


1.74929 


5.53173 


3.66 


13.3956 


1.91311 


6.04979 


4.26 


18.1476 


2.06398 


6.52687 


3.07 


9.4249 


1.75214 


5.54076 


3.67 


13.4689 


1.91572 


6.05805 


4.27 


18.2329 


2.06640 


6.53452 


3.08 


9.4864 


1.75499 


5.54977 


3.68 


13.5424 


1.91833 


6.06630 


4.28 


18.3184 


2.06882 


6.54217 


3.09 


9.5481 


1.75784 


5.55878 


3.69 


13.6161 


1.92094 


6.07454 


4.29 


18.4041 


2.07123 


6.54981 


3.10 

3.11 


9.6100 


1.76068 


5.56776 


3.70 

3.71 


13.6900 


1.92354 


6.08276 


4.80 

4.31 


18.4900 


2.07364 


6.55744 


9.6721 


1.70352 


5.57674 


13.7641 


1.92614 


6.09098 


18.5761 


2.07605 


6.56506 


3.12 


9.7344 


1.76635 


5.58570 


3.72 


13.8384 


1.92873 


6.09918 


4.32 


18.6624 


2.07846 


6.57267 


3.13 


9.7969 


1.76918 


5.59464 


3.73 


13.9129 


1.93132 


6.10737 


4.33 


18.7489 


2.08087 


6.58027 


3.14 


9.8596 


1.77200 


5.60357 


3.74 


13.9876 


1.93391 


6.11555 


4.34 


18.8356 


2.08327 


6.58787 


3.15 


9.9225 


1.77482 


5.61249 


8.75 


14.0625 


1.93649 


6.12372 


4.35 


18.9225 


2.08567 


6.59545 


3.16 


9.9856 


1.77764 


5.62139 


3.76 


14.1376 


1.93907 


6.13188 


4.36 


19.0096 


2.08806 


6.60303 


3.17 


10.0489 


1.78045 


5.63028 


3.77 


14.2129 


1.94165 


6.14003 


4.37 


19.0969 


2.09045 


6.61060 


3.18 


10.1124 


1.78326 


5.63915 


3.78 


14.2884 


1.94422 


6.14817 


4.38 


19.1844 


2.09284 


6.61816 


8.19 


10.1761 


1.78606 


5.64801 


3.79 


14.3641 


1.94679 


6.15630 


4.39 


19.2721 


2.09523 


6.62671 


3.10 

3.21 


10.2400 


1.78885 


5.65685 


3.80 

3.81 


14.4400 


1.94936 


6.16441 


4.40 

4.41 


19.3600 


2.09762 


6.63325 


10.3041 


1.79165 


5.66569 


14.5161 


1.95192 


6.17252 


19.4481 


2.10000 


6.64078 


3.22 


10.3684 


1.79444 


5.67450 


3.82 


14.5924 


1.95448 


6.18061 


4.42 


19.5364 


2.10238 


6.64831 


3.23 


10.4329 


1.79722 


5.68331 


3.83 


14.6689 


1.95704 


6.18870 


4.43 


19.6249 


2.10476 


6.65582 


3.24 


10.4976 


1.80000 


5.69210 


3.84 


14.7456 


1.95959 


6.19677 


4.44 


19.7136 


2.10713 


6.66333 


3.15 


10.5625 


1.80278 


5.70088 


3.85 


14.8225 


1.96214 


6.20484 


4.46 


19.8025 


2.10950 


6.67083 


3.26 


10.6276 


1.80555 


5.70964 


3.86 


14.8996 


1.96469 


6.21289 


4.46 


19.8916 


2.11187 


6.67832 


3.27 


10.6929 


1.80S31 


5.71839 


3.87 


14.9769 


1.96723 


6.22093 


4.47 


19.9809 


2.11424 


6.68581 


3.28 


10.7584 


1.81108 


5.72713 


3.88 


15.0544 


1.96977 


6.22896 


4.48 


20.0704 


2.11660 


6.69328 


3.29 


10.8241 


1.81384 


5.73585 


3.89 


15.1321 


1.97231 


6.23699 


4.49 


20.1601 


2.11896 


6.70075 


3.30 
3.31 


10.8900 


1.81659 


5.74456 


3.90 

3.91 


15.2100 


1.97484 


6.24500 


4.60 
4.51 


20.2500 


2.12132 


6.70820 


10.9561 


1.81934 


5.75326 


15.2881 


1.97737 


6.25300 


20.3401 


2.12368 


6.71565 


3.32 


11.0224 


1.82209 


5.76194 


3.92 


15.3664 


1.97990 


6.26099 


4.52 


20.4304 


2.12603 


6.72309 


3.33 


11.0889 


1.82483 


5.77062 


3.93 


15.4449 


1.98242 


6.26897 


4.53 


20.5209 


2.12838 


6.73053 


3.34 


11.1556 


1.82757 


5.77927 


3.94 


15.5236 


1.98494 


6.27694 


4.54 


20.6116 


2.13073 


6.73795 


8.35 


11.2225 


1.83030 


5.78792 


3.96 


15.6025 


1.98746 


6.28490 


4.55 


20.7025 


2.13307 


6.74537 


3.36 


11.2896 


1.83303 


5.79655 


3.96 


15.6816 


1.98997 


6.29285 


4.56 


20.7936 


2.13542 


6.75278 


3.37 


113569 


1.83576 


5.80517 


3.97 


15.7609 


1.99249 


6.30079 


4.57 


20.8849 


2.13776 


6.76018 


3.38 


11.4244 


1.83848 


5.81378 


3.98 


15.8404 


1.99499 


6.30872 


4.58 


20.9764 


2.14009 


6.76757 


3.39 


11.4921 


1.84120 


5.82237 


3.99 


15.9201 


1.99750 


6.31664 


4.59 


21.0681 


2.14243 


6.77495 


8.40 


11.5600 


1.84391 


5.83095 


4.00 

N 


16.0000 


2.00000 


6.32456 


4.60 


21.1600 


2.14476 


6.78233 


N 


N* 


Vn 


Vxofi 


N* 


Vn 


Viw 


S 


iV» 


V* 


Vmr 



334 



Appendix A 











TABLE V {continued) 










N 
4.60 


N* 


Vn 

2.14476 


VlOAT 


N 


N* 


Vn 

2.28035 


Vion 


N 


N* 


Vn 


VlON 


21.1600 


6.78233 


5 JO 


27.0400 


7.21110 


5.80 


33.6400 


2.40832 


7.61577 


4.61 


21.2521 


2.14709 


6.78970 


5.21 


27.1441 


2.28254 


7.21803 


5.81 


33.7561 


2.41039 


7.62234 


4.62 


21.3444 


2.14942 


6.79706 


5.22 


27.2484 


2.28473 


7.22496 


5.82 


33.8724 


2.41247 


7.62889 


4.03 


21.4369 


2.15174 


6.80441 


5.23 


27.3529 


2.28692 


7.23187 


5.83 


33.9889 


2.41454 


7.63544 


4.64 


21.5296 


2:15407 


6.81175 


5.24 


27.4576 


2.28910 


7.23878 


5.84 


34.1056 


2.41661 


7.64199 


4.66 


21.6225 


2.15639 


6.81909 


5.25 


27.5625 


2.29129 


7.24569 


5.85 


34.2225 


2.41868 


7.64853 


4.66 


21.7156 


2.15870 


6.82642 


5.26 


27.6676 


2.29347 


7.25259 


5.86 


34.3396 


2.42074 


7.65506 


4.67 


21.8089 


2.16102 


6.83374 


5.27 


27.7729 


2.29565 


7.25948 


5.87 


34.4569 


2.42281 


7.66159 


4.68 


21.9024 


2.16333 


6.84105 


5.28 


27.8784 


2.29783 


7.26636 


5.88 


34.5744 


2.42487 


7.66812 


4.69 


21.9961 


2.16564 


6.84836 


5.29 


27.9841 


2.30000 


7.27324 


5.89 


34.6921 


2.42693 


7.67463 


4.70 

4.71 


22.0900 


2.16795 


6.85565 


5.30 

5.31 


28.0900 


2.30217 


7.28011 


5.90 

5.91 


34.8100 


2.42899 


7.68115 
7.68765 


22.1841 


2.17025 


6.86294 


28.1961 


2.30434 


7.28697 


34.9281 


2.43105 


4.72 


22.2784 


2.17256 


6.87023 


5.32 


28.3024 


2.30651 


7.29383 


5.92 


35.0464 


2.43311 


7.69415 


4.73 


22.3729 


2.17486 


6.87750 


5.33 


28.4089 


2.30868 


7.30068 


5.93 


35.1649 


2.43516 


7.70065 


4.74 


22.4676 


2.17715 


6.88477 


5.34 


28.5156 


2.31084 


7.30753 


5.94 


35.2836 


2.43721 


7.70714 


4.76 


22.5625 


2.17945 


6.89202 


5.35 


28.6225 


2.31301 


7.31437 


5.95 


35.4025 


2.43926 


7.71362 


4.76 


22.6576 


2.18174 


6.89928 


5.36 


28.7296 


2.31517 


7.32120 


5.96 


35.5216 


2.44131 


7.72010 


4.77 


?2.7529 


2.18403 


6.90652 


5.37 


28.8369 


2.31733 


7.32803 


5.97 


35.6409 


2.44336 


7.72658 


4.78 


22.8484 


2.18632 


6.91375 


5.38 


28.9444 


2.31948 


7.33485 


5.98 


35.7604 


2.44540 


7.73305 


4.79 


22.9441 


2.18861 


6.92098 


5.39 


29.0521 


2.32164 


7.34166 


5.99 


35.8801 


2.44745 


7.73951 


4.80 

4.81 


23.0400 
23.136? 


2.19089 


6.92820 


5.40 

5.41 


29.1600 


2.32379 


7.34847 


6.00 

6.01 


36.0000 


2.44949 


7.74597 


2.19317 


6.93542 


29.2681 


2.32594 


7.35527 


36.1201 


2.45153 


7.75242 


4.82 


23.2324 


2.19545 


6.94262 


5.42 


29.3764 


2.32809 


7.36206 


6.02 


36.2404 


2.45357 


7.75887 


4.83 


23.3289 


2.19773 


6.94982 


5.43 


29.4849 


2.33024 


7.36885 


6.03 


36.3609 


2.45561 


7.76531 


4.84 


23.4256 


2.20000 


6.95701 


5.44 


29.5936 


2.33238 


7.37564 


6.04 


36.4816 


2.45764 


7.77174 


4.86 


23.5225 


2.20227 


6.96419 


5.45 


29.7025 


2.33452 


7.38241 


6.05 


36.6025 


2.45967 


7.77817 


4.86 


23.6196 


2.20454 


6.97137 


5.46 


29.8116 


2.33666 


7.38918 


6.06 


36.7236 


2.46171 


7.78460 


4.87 


23.7169 


2.20681 


6.97854 


5.47 


29.9209 


2.33880 


7.39594 


6.07 


36.8449 


2.46374 


7.79102 


4.88 


23.8144 


2.20907 


6.98570 


5.48 


30.0304 


2.34094 


7.40270 


6.08 


36.9664 


2.46577 


7.79744 


4.89 


23.9121 


2.21133 


6.99285 


5.49 


30.1401 


2.34307 


7.40945 


6.09 


37.0881 


2.46779 


7.80385 


4.90 

4.91 


24.0100 


2.21359 


7.00000 


5.60 

5.51 


30.2500 


2.34521 


7.41620 


6.10 

6.11 


37.2100 


2.46982 


7.81025 


24.1081 


2.21585 


7.00714 


30.3601 


2.34734 


7.42294 


37.3321 


2.47184 


7.81665 


4.92 


24.2064 


2.21811 


7.01427 


5.52 


30.4704 


2.34947 


7.42967 


6.12 


37.4544 


2.47386 


7.82304 


4.93 


24.3049 


2.22036 


7.02140 


5.53 


30.5809 


2.35160 


7.43640 


6.13 


37.5769 


2.47588 


7.82943 


4.94 


24.4036 


2.22261 


7.02851 


5.54 


30.6916 


2.35372 


7.44312 


6.14 


37.6996 


2.47790 


7.83582 


4.95 


24.5025 


2.22486 


7.03562 


5.55 


30.8025 


2.35584 


7.44983 


6.16 


37.8225 


2.47992 


7.84219 


4.96 


24.6016 


2.22711 


7.04273 


5.56 


30.9136 


2.35797 


7.45654 


6.16 


37.9456 


2.48193 


7.84857 


4.97 


24.7009 


2.22935 


7.04982 


5.57 


31.0249 


2.36008 


7.46324 


6.17 


38.0689 


2.48395 


7.85493 


4.98 


24.8004 


2.23159 


7.05691 


5.58 


31.1364 


2.36220 


7.46994 


6.18 


38.1924 


2.48596 


7.86130 


4.99 


24.9001 


2.23383 


7.06399 


5.59 


31.2481 


2.36432 


7.47663 


6.19 


38.3161 


2.48797 


7.86766 


6.00 

5.01 


25.0000 


2.23607 


7.07107 


5.60 

5.61 


31.3600 


2.36643 


7.48331 


6 JO 

6.21 


38.4400 


2.48998 


7.87401 


25.1001 


2.23830 


7.07814 


31.4721 


2.36854 


7.48999 


38.5641 


2.49199 


7.88036 


5.02 


25.2004 


2.24054 


7.08520 


5.62 


31.5844 


2.37065 


7.49667 


6.22 


38.6884 


2.49399 


7.88670 


5.03 


25.3009 


2.24277 


7.09225 


5.63 


31.6969 


2.37276 


7.50333 


6.23 


38.8129 


2.49600 


7.89303 


5.04 


25.4016 


2.24499 


7.09930 


5.64 


31.8096 


2.37487 


7.50999 


6.24 


38.9376 


2.49800 


7.89937 


5.06 


25.5025 


2.24722 


7.10634 


5.66 


31.9225 


2.37697 


7.51665 


6.25 


39.0625 


2.50000 


7.90569 


5.06 


25.6036 


2.24944 


7.11337 


5.66 


32.0356 


2.37908 


7.52330 


6.26 


39.1876 


2.50200 


7.91202 


5.07 


25.7049 


2.25167 


7.12039 


5.67 


32.1489 


2.38118 


7.52994 


6.27 


39.3129 


2.50400 


7.91833 


5.08 


25.8064 


2.25389 


7.12741 


5.68 


32.2624 


2.38328 


7.53658 


6.28 


39.4384 


2.50599 


7.92465 


5.09 


25.9081 


2.25610 


7.13442 


5.69 


32.3761 


2.38537 


7.54321 


6.29 


39.5641 


2.50799 


7.93095 


6.10 

5.11 


26.0100 


2.25832 


7.14143 


6.70 

5.71 


32.4900 


2.38747 


7.54983 


6.30 

6.31 


39.6900 


2.50998 


7.93725 


26.1121 


2.26053 


7.14843 


32.6041 


2.38956 


7.55645 


39.8161 


2.51197 


7.94355 


5.12 


26.2144 


2.26274 


7.15542 


5.72 


32.7184 


2.39165 


7.56307 


6.32 


39.9424 


2.51396 


7.94984 


5.13 


26.3169 


2.26495 


7.16240 


5.73 


32.8329 


2.39374 


7.56968 


6.33 


40.0689 


2.51595 


7.95613 


5.14 


26.4196 


2.26716 


7.16938 


5.74 


32.9476 


2.39583 


7.57628 


6.34 


40.1956 


2.51794 


7.96241 


5.16 


26.5225 


2.26936 


7.17635 


5.75 


33.0625 


2.39792 


7.58288 


6.35 


40.3225 


2.51992 


7.96869 


5.16 


26.6256 


2.27156 


7.18331 


5.76 


33.1776 


2.40000 


7.58947 


6.36 


40.4496 


2.52190 


7.97496 


5.17 


26.7289 


2.27376 


7.19027 


5.77 


33.2929 


2.40208 


7.59605 


6.37 


40.5769 


2.52389 


7.98123 


5.18 


26.8324 


2.27596 


7.19722 


$.78 


33.4084 


2.40416 


7.60263 


6.38 


40.7044 


2.52587 


7.98749 


5.19 


26.9361 


2.27816 


7.20417 


5.79 


33.5241 


2.40624 


7.60920 


6.39 


40.8321 


2.52784 


7.99375 


5.20 

N 


27.0400 


2.28035 


7.21110 


5.60 


33.6400 


2.40832 


7.61577 


6.40 


40.9600 


2.52982 


8.00000 


AP 


Vn 


VTqn 


N 


iV» 


Vn 


VTqn 


N 


N* 


Vn 


vT<w 



Appendix A 



335 











TABLE V (cat 


itinued] 


) 








N 


N* 


Vn 


VlON 


N 

7.00 


N* 


Vn 


VlON 


N 

7.60 


JV» 


Vn 


Vwf 


6.40 


40.9600 


2.52982 


8.00000 


49.0000 


2.64575 


8.36660 


57.7600 


2.75681 


8.71780 


6.41 


41.0881 


2.53180 


8.00625 


7.01 


49.1401 


2.64764 


8.37257 


7.61 


57.9121 


2.75862 


8.72353 


6.42 


41.2164 


2.53377 


8.01249 


7.02 


49.2804 


2.64953 


8.37854 


7.62 


58.0644 


2.76043 


8.72926 


0.43 


41.3449 


2.53574 


8.01873 


7.03 


49.4209 


2.65141 


8.38451 


7.63 


58.2169 


2.76225 


8.73499 


6.44 


41.4736 


2.53772 


8.02496 


7.04 


49.5616 


2.65330 


8.39047 


7.64 


58.3696 


2.76405 


8.74071 


•.49 


41.6025 


2.53969 


8.03119 


7.06 


49.7025 


2.65518 


8.39643 


7.66 


58.6225 


2.76586 


8.74643 


6.46 


41.7316 


2.54165 


8.03741 


7.06 


49.8436 


2.65707 


8.40238 


7.66 


58.6756 


2.76767 


8.75214 


6.47 


41.8609 


2.54362 


8.04363 


7.07 


49.9849 


2.65895 


8.40833 


7.67 


58.8289 


2.76948 


8.75785 


6.48 


41.9904 


2.54558 


8.04984 


7.08 


50.1264 


2.66083 


8.41427 


7.68 


58.9824 


2.77128 


8.76356 


A.49 


42.1201 


2.54755 


8.05605 


7.09 


50,2681 


2.66271 


8.42021 


7.69 


59.1361 


2.77308 


8.76926 


6.50 

6.51 


42.2500 


2.54951 


8.06226 


7.10 

7.11 


50.4100 


2.66458 


8.42615 


7.70 

7.71 


59.2900 


2.77489 


8.77496 


42.3801 


2.55147 


8.06846 


50.5521 


2.66646 


8.4321)8 


59.4441 


2.77669 


8.78066 


6.52 


42.5104 


2.55343 


8.07465 


7.12 


50.6944 


2.66833 


8.43801 


7.72 


59.6984 


2.77849 


8.78635 


6.53 


42.6409 


2.55539 


8.08084 


7.13 


50.8369 


2.67021 


8.44393 


7.73 


69.7529 


2.78029 


8.79204 


6.54 


42.7716 


2.55734 


8.06703 


7.14 


60.9796 


2.67208 


8.44985 


7.74 


59.9076 


2.78209 


8.79773 


6.56 


42.9025 


2.55930 


8.09321 


7.15 


51.1225 


2.67395 


8.45577 


7.75 


60.0625 


2.78388 


8.80341 


6.56 


43.0336 


2.56125 


8.09938 


7.16 


51.2656 


2.67582 


8.46168 


7.76 


60.2176 


2.78568 


8.80909 


6.57 


43.1649 


2.56320 


8.10555 


7.17 


51.4089 


2.67769 


8.46759 


7.77 


60.3729 


2.78747 


8.81476 


6.58 


43.2964 


2.56515 


8.11172 


7.18 


51.5524 


2.67955 


8.47349 


7.78 


60.5284 


2.78927 


8.82043 


6.59 


43.4281 


2.56710 


8.11788 


7.19 


51.6961 


2.68142 


8.47939 


7.79 


60.6841 


2.79106 


8.82610 


6.60 

6.61 


43.5600 


2.56905 


8.12404 


7 JO 

7.21 


51.8400 


2.68328 


8.48528 


7.80 

7.81 


60.8400 


2.79285 


8.83176 


43.6921 


2.57099 


8.13019 


51.9841 


2.68614 


8.49117 


60.9961 


2.79464 


8.83742 


6.62 


43.8244 


2.57294 


8.13634 


7.22 


52.1284 


2.68701 


8.49706 


7.82 


61.1524 


2.79643 


8.84308 


6.63 


43.9569 


2.57488 


8.14248 


7.23 


52.2729 


2.68887 


8.50294 


7.83 


61.3089 


2.79821 


8.84873 


6.64 


44.0896 


2.57682 


8.14862 


7.24 


52.4176 


2.69072 


8.50882 


7.84 


61.4656 


2.80000 


8.85438 


6.M 


44.22% 


2.57876 


8.15475 


7J5 


52.5625 


2.69258 


8.51469 


7J6 


61.6225 


2.80179 


8.86002 


6.66 


44.3556 


2.58070 


8.16088 


7.26 


52.7076 


2.69444 


8.52056 


7.86 


61.7796 


2.80357 


8.86566 
8.87130 


6.67 


44.4889 


2.58263 


8.16701 


7.27 


52.8529 


2.69629 


8.62643 


7.87 


61.9369 


2.80535 


6.68 


44.6224 


2.58457 


8.17313 


7.28 


52.9984 


2.69815 


8.53229 


7.88 


62.0944 


2.80713 


8.87694 


6.69 


44.7561 


2.58650 


8.17924 


7.29 


53.1441 


2.70000 


8.53815 


7.89 


62.2521 


2.80891 


8.88257 


6.70 

6.71 


44.8900 


2.58844 


8.18535 


7.30 

7.31 


53.2900 


2.70185 


8.54400 


7.90 

7.91 


62.4100 


2.81069 


8.88819 


45.0241 


2.59037 


8.19146 


53.4361 


2.70370 


8.54985 


62.5681 


2.81247 


8.89382 


6.72 


45.1584 


2.59230 


8.19756 


7.32 


53.5824 


2.70555 


8.55570 


7.92 


62.7264 


2.81425 


8.89944 


6.73 


45.2929 


2.59422 


8.20366 


7.33 


53.7289 


2.70740 


8.56154 


7.93 


62.8849 


2.81603 


8.90505 


6.74 


45.4276 


2.59615 


8.20975 


7.34 


53.8756 


2.70924 


8.56738 


7.94 


63.0436 


2.81780 


8.91067 


►6.70 


45.5625 


2.59808 


8.21584 


7.36 


54.0225 


2.71109 


8.57321 


7.96 


63.2025 


2.81957 


8.91628 


6.76 


45.6976 


2.60000 


8.22192 


7.36 


54.1696 


2.71293 


8.57904 


7.96 


63.3616 


2.82135 


8.92188 


6.77 


45.8329 


2.60192 


8.22800 


7.37 


54.3169 


2.71477 


8.58487 


7.97 


63.5209 


2.82312 


8.92749 


6.78 


45.9684 


2.60384 


8.23408 


7.38 


54.4644 


2.71662 


8.59069 


7.98 


63.6804 


2.82489 


8.93308 


6.79 


46.1041 


2.60576 


8.24015 


7.39 


54.6121 


2.71846 


8.59651 


7.99 


63.8401 


2.82666 


8.93868 


6.80 

6.81 


46.2400 


2.60768 


8.24621 


7.40 

7.41 


54.760Q 


2.72029 


8.60233 


8.00 

8.01 


64.0000 


2.82843 


8.94427 


46.3761 


2.60960 


8.25227 


54.9081 


2.72213 


8.60814 


64.1601 


2.83019 


8.94986 


6.82 


46.5124 


2.61151 


8.25833 


7.42 


55.0564 


2.72397 


8.61394 


8.02 


64.3204 


2.83196 
2.83373 


8.95545 


6.83 


46.6489 


2.61343 


8.26438 


7.43 


55.2049 


2.72580 


8.61974 


8.03 


64.4809 


8.96103 


6.84 


46.7856 


2.61534 


8.27043 


7.44 


55.3536 


2.72764 


8.62554 


8.04 


64.6416 


2.83549 


8.96660 


6.86 


46.9225 


2.61725 


8.27647 


7.46 


55.5025 


2.72947 


8.63134 


8.06 


64.8025 


2.83725 


8.97218 


6.86 


47.0596 


2.61916 


8.28251 


7.46 


55.6516 


2.73130 


8.63713 


8.06 


64.9636 


2.83901 


8.97775 


6.87 


47.1969 


2.62107 


8.28855 


7.47 


55.8009 


2.73313 


8.64292 


8.07 


65.1249 


2.84077 


8.98332 


6.88 


47.3344 


2.62208 


8.29458 


7.48 


55.9504 


2.73496 


8.64870 


8.08 


65.2864 


2.84253 


8.98888 


6.89 


47.4721 


2.62488 


8.30060 


7.49 


56.1001 


2.73679 


8.65448 


8.09 


65.4481 


2.84429 


8.99444 


6.90 

6.91 


47.6100 


2.62679 


8.30662 


7.60 

7.51 


56.2500 


2.73861 


8.66025 


8.10 

8.11 


65.6100 


2.84605 


9.00000 


47.7481 


2.62869 


8.31264 


56.4001 


2.74044 


8.66603 


65.7721 


2.84781 


9.00555 


6.92 


47.8864 


2.63059 


8.31865 


7.52 


56.5504 


2.74226 


8.67179 


8.12 


65.9344 


2.84956 


9.01110 


6.93 


48.0249 


2.63249 


8.32466 


7.53 


56.7009 


2.74408 


8.67756 


8.13 


66.0969 


2.85132 


9.01665 


6.94 


48.1636 


2.63439 


8.33067 


7.54 


56.8516 


2.74591 


8.68332 


8.14 


66.2596 


2.85307 


9.02219 


6.95 


48.3025 


2.63629 


8.33667 


7.66 


67.0025 


2.74773 


8.68907 


8.16 


66.4225 


2.85482 


9.02774 


6.96 


48.4416 


2.63818 


8.34266 


7.56 


57.1536 


2.74955 


8.69483 


8.16 


66.5856 


2.85657 


9.03327 


6.97 


48.5809 


2.64008 


8.34865 


7.57 


57.3049 


2.75136 


8.70057 


8.17 


66.7489 


2.85832 


9.03881 


6.98 


48.7204 


2.64197 


8.35464 


7.58 


57.4564 


2.75318 


8.70632 


8.18 


66.9124 


2.86007 


9.04434 


6.99 


48.8601 


2.64386 


8.36062 


7.59 


57.6081 


2.75500 


8.71206 


8.19 


67.0761 


2.86182 


9.04986 


7.00 

N 


49 0000 


2.64575 


8.36660 


7.60 
N 


57.7600 


2.75681 


8.71780 


8.20 

AT 


67.2400 


2.86356 


9.05539 


N* 


VH 


Vudi 


N* 


Vn 


van 


N* 


Vn 


V$5R 



336 



Appendix A 











TABLE V (con 


tinued) 










H 


N* 


Vn 


viw? 


N 


JV« 


Vn 


vISat 


N 


N* 


Vii 


Vufir 


8 JO 


67.2400 


2.86356 


9.05539 


8.80 


77.4400 


2.96648 


9.38083 


9.40 


88.3600 


3.06594 


9.69536 


8.21 


67.4041 


2.86531 


9.06091 


8.81 


77.6161 


2.96816 


9.38616 


9.41 


88.5481 


3.06757 


9.70052 


8.22 


67.5684 


2.86705 


9.06642 


8.82 


77.7924 


2.96985 


9.39149 


9.42 


88.7364 


3.06920 


9.70567 


8.23 


67.7329 


2.86880 


9.07193 


8.83 


77.9689 


2.97153 


9.39681 


9.43 


88.9249 


3.07083 


9.71082 


8.24 


67.8976 


2.87054 


9.07744 


8.84 


78.1456 


2.97321 


9.40213 


9.44 


89.1136 


3.07246 


9.71597 


8J5 


68.0625 


2.87228 


908295 


8.86 


78.3225 


2.97489 


9.40744 


9.45 


89.3025 


3.07409 


9.72111 


8.26 


68.2276 


2.87402 


9.08845 


8.86 


78.4996 


2.97658 


9.41276 


9.46 


89.4916 


3.07571 


9.72625 


8.27 


68.3929 


2.87576 


9.09395 


8.87 


78.6769 


2.97826 


9.41807 


9.47 


89.6809 


3.07734 


9.73139 


8.28 


68.5584 


2.87750 


9.09945 


8.88 


78.8544 


2.97993 


9.42338 


9.48 


89.8704 


3.07896 


9.73653 


8.29 


68.7241 


2.87924 


9.10494 


8.89 


79.0321 


2.98161 


9.42868 


9.49 


90.0601 


3.08058 


9.74166 


8.80 

8.31 


68.8900 


2.88097 


9.11043 


8.90 

8.91 


79.2100 


2.98329 


9.43398 


9.50 

9.51 


90.2500 


3.08221 


9.74679 


69.0561 


2.88271 


9.11592 


79.3881 


2.98496 


9.43928 


90.4401 


3.08383 


9.75192 


8.32 


69.2224 


2.88444 


$.12140 


8.92 


79.5664 


2.98664 


9.44458 


9.52 


90.6304 


3.08545 


9.75705 


8.33 


69.3889 


2.88617 


9.12688 


8.93 


79.7449 


2.98831 


9.44987 


9.53 


90.8209 


3.08707 


9.76217 


8.34 


69.5556 


2.88791 


9.13236 


8.94 


79.9236 


2.98998 


9.45516 


9.54 


91.0116 


3.08869 


9.76729 


8J6 


69.7225 


2.88964 


9.13783 


8.95 


80.1025 


2.99166 


9.46044 


9.65 


91.2025 


3.09031 


9.77241 


8.36 


69.8896 


2.89137 


9.14330 


8.96 


80.2816 


2.99333 


9.46573 


9.56 


91.3936 


3.09192 


9.77753 


8.37 


70.0569 


2.89310 


9.14877 


8.97 


80.4609 


2.99500 


9.47101 


9.67 


91.5849 


3.09354 


9.78264 


8.38 


70.2244 


2.89482 


9.15423 


8.98 


80.6404 


2.99666 


9.47629 


9.58 


91.7764 


3.09516 


9.78775 


8.39 


70.3921 


2.89655 


9.15969 


8.99 


80.8201 


2.99833 


9.48156 


9.59 


91.9681 


3.09677 


9.79285 


8.40 

8.41 


70.5600 


2.89828 


9.16515 


9.00 

9.01 


81.0000 


3.00000 


9.48683 


9.60 

9.61 


92.1600 


3.09839 


9.79796 


70.7281 


2.90000 


9.17061 


81.1801 


3.00167 


9.49210 


92.3521 


3.10000 


9.80306 


8.42 


70.8964 


2.90172 


9.17606 


9.02 


81.3604 


3.00333 


9.49737 


9.62 


92.5444 


3.10161 


9.80816 


8.43 


71.0649 


2.90345 


9.18150 


9.03 


81.5409 


3.00500 


9.50263 


9.63 


92.7369 


3.10322 


9.81326 


8.44 


71.2336 


2.90517 


9.18695 


9.04 


81.7216 


3.00666 


9.50789 


9.64 


92.9296 


3.10483 


9.81835 


8.45 


71.4025 


2.90689 


9.19239 


9.05 


81.9025 


3.00832 


9.51315 


9.65 


93.1225 


3.10644 


9.82344 


8.46 


71.5716 


2.90861 


9.19783 


9.06 


82.0836 


3.00998 


9.51840 


9.66 


93.3156 


3.10805 


9.82853 


8.47 


71.7409 


2.91033 


9.20326 


9.07 


82.2649 


3.01164 


9.52365 


9.67 


93.5089 


3.10966 


9.83362 


8.48 


71.9104 


2.91204 


9.20869 


9.08 


82.4464 


3.01330 


9.52890 


9.68 


93.7024 


3.11127 


9.83870 


8.49 


72.0801 


2.91376 


9.21412 


9.09 


82.6281 


3.01496 


9.53415 


9.69 


93.8961 


3.11288 


9.84378 


8.60 

8.51 


72.2500 


2.91548 


9.21954 


9.10 

9.11 


82.8100 


3.01662 


9.53939 


9.70 

9.71 


94.0900 


3.11448 


9.84886 


72.4201 


2.91719 


9.22497 


82.9921 


3.01828 


9.54463 


94.2841 


3.11609 


9.85393 


8.52 


72.5904 


2.91890 


9.23033 


9.12 


83.1744 


3.01993 


9.54987 


9.72 


94.4784 


3.11769 


9.85901 


8.53 


72.7609 


2.92062 


9.23580 


9.13 


83.3569 


3.02159 


9.55510 


9.73 


94.6729 


3.11929 


9.86408 


8.54 


72.9316 


2.92233 


9.24121 


9.14 


83.5396 


3.02324 


9.56033 


9.74 


94.8676 


3.12090 


9.86914 


8.65 


73.1025 


2.92404 


9.24662 


9.15 


83.7225 


3.02490 


9.56556 


9.76 


95.0625 


3.12250 


9.87421 


8.56 


73.2736 


2.92575 


9.25203 


9.16 


83.9056 


3.02655 


9.57079 


9.76 


95.2576 


3.12410 


9.87927 


8.57 


73.4449 


2.92746 


9.25743 


9.17 


84.0889 


3 02820 


9.57601 


9.77 


95.4529 


3.12570 


9.88433 


8.58 


73.6164 


2.92916 


9.26283 


9.18 


84.2724 


3.02985 


9.58123 


9.78 


95.6484 


3.12730 


9.88939 


8.59 


73.7881 


2.93087 


9.26823 


9.19 


84.4561 


3.03150 


9.58645 


9.79 


95.8441 


3.12890 


9.89444 


8.80 

8.61 


73.9600 


2.93258 


9.27362 


9 JO 

9.21 


84.6400 


3.03315 


9.59166 


9.80 

9.81 


96.0400 


3.13050 


9.89949 


74.1321 


2.93428 


9.27901 


84.8241 


3.03480 


9.59687 


96.2361 


3.13209 


9.90454 


8.62 


74.3044 


2.9359S 


9.23440 


9.22 


85.0084 


3.03645 


9.60208 


9.82 


96.4324 


3.13369 


9.90959 


8.63 


74.4769 


2.93769 


9.28978 


9.23 


85.1929 


3.03809 


9.60729 


9.83 


96.6289 


3.13528 


9.91464 


8.64 


74.6496 


2.93939 


9.29516 


9.24 


85.3776 


3.03974 


9.61249 


9.84 


96.8256 


3.13688 


9.91968 


8.68 


74.8225 


2.94109 


9.30054 


9J5 


85.5625 


3.04138 


9.61769 


9.85 


97.0225 


3.13847 


9.92472 


8.66 


74.9956 


2.94279 


9.30591 


9.26 


85.7476 


3.04302 


9.62289 


9.86 


97.2196 


3.14006 


9.92975 


8.67 


75.1689 


2.94449 


9.31128 


9.27 


85.9329 


3.04467 


9.62808 


9.87 


97.4169 


3.14166 


9.93479 


8.68 


75.3424 


2.94618 


9.31665 


9.28 


861184 


3.04631 


9.63328 


9.88 


97.6144 


3.14325 


9.93982 


8.69 


75.5161 


2.94788 


9.32202 


9.29 


86.3041 


3.04795 


9.63846 


9.89 


97.8121 


3.14484 


9.94485 


8.70 

8.71 


75.6900 


2.94958 


9.32738 


9.80 

9.31 


86.4900 


3.04959 


9.64365 


9.90 

9.91 


98.0100 


3.14643 


9.94987 


75.8641 


2.95127 


9.33274 


86.6761 


3.05123 


9.64883 


98.2081 


3.14802 


9.95490 


8.72 


76.0384 


2.95296 


9.33809 


9.32 


86.8624 


3.05287 


9.65401 


9.92 


98.4064 


3.14960 


9.95992 


8.73 


76.2129 


2.95466 


9.34345 


9.33 


87.0489 


3.05450 


9.65919 


9.93 


98.6049 


3.15119 


9.96494 


8.74 


76.3876 


2.95635 


9.34880 


9.34 


87.2356 


3.05614 


9.66437 


9.94 


98.8036 


3.15278 


9.96995 


8.75 


76.5625 


2.95804 


9.35414 


9.35 


87.4225 


3.05778 


9.66954 


9.95 


99.0025 


3.15436 


9.97497 


8.76 


76.7376 


2.95973 


9.35949 


9.36 


87.6096 


3.05941 


9.67471 


9.96 


99.2016 


3.15595 


9.97998 


8.77 


76.9129 


2.96142 


9.36483 


9.37 


87.7969 


3.06105 


9.67988 


9.97 


99.4009 


3.15753 


9.98499 


8.78 


77.0884 


2.96311 


9.37017 


9.38 


87.9844 


3.06268 


9.68504 


9.98 


99.6004 


3.15911 


9.98999 


8.79 


77.2641 


2.96479 


9.37550 


9.39 


88.1721 


3.06431 


9.69020 


9.99 


99.8001 


3.16070 


9.99500 


8.80 

* 


77.4400 


2.96648 


9.38083 


9.40 
N 


88.3600 


3.06594 


9.69536 


10.00 


100.000 


3.16228 


10.0000 


N* VN 


Vtoii 


N* 


VN 


VlON 


N 


N* 


VN 


Vioif 



EXERCISE 1-1. PAGE 11 

1. (a) Commutative law of addition. 
(c) Associative law of multiplication. 
(e) Commutative law of multiplication. 

2. (a) 2. (c) 1. (e) 1. (g) -5. (i) -2. (k) -7. 

3. (a) -6. (c) -35. (e) -10. (g) 2. (i) 29. 

4. (a) -5. (c) 0. (e) -2/3. (g) 36 - 2a. (i) -a6. (k) 3 - x. 

5. (a)l. (c)5/17. (e) 1/1.02. <*> -™- <S)x+V- (k) r - 0.1. 

EXERCISE 1-2. PAGE 15 

1. (a) -3, -2, 0, 4, 5. _(c) -4, -2, -1, -1/3, 10. (e) -2, -3/2, 1, y/3, 3. 

2. (a) 3 > 1/3. (c) V2 > 1.414. (e) 22/7 > it. 

3. (a) False, (c) True, (e) False. 

4. (a) 0. (c) 0. (e) 2. (g) ^6. (i) 0. 

5. (a) -1 < x < 1. (c) -a £ x ^ a. (e) -2 < x < 4. 

6. (a) 4 < 5 < 6. (c) -1 < < 1. (e) 1 < y/Z < 2. 

7. a + 7 ^ 10. 

EXERCISE 1-3. PAGE 20 

1.32. 3.-1. 5.1,000,000. 7.-216. 9.2401. 11. —• 13. 16 ' 384 



16 823,543 

15.^- 17.32. 19.125a. 21. a\ 23. a 4 6 4 . 25. a 4 6°. 27. a". 

29. a 2 »6 2 ™. 31. a 4 . 33. -• 35. —• 37. 5 - 6. 39. 46. 41. 3a - 46. 

a 4 a 4 

43. -2a - 36. 45. (a - 2c)x 2 - (a + 26)*?/ + (6 - l)y 2 . 47. 4a - 56 - 2c. 

49. a + (6 + c), a - (-6 - c). 51. a 2 + (c 2 - 6 2 ), a 2 - (6 2 - c 2 ). 

53. 2a + (6 - 3c), 2a - (3c - 6). 55. x 2 + (- y 2 - z 2 ), x 2 - (y 2 + z 2 ). 

57. -a 3 6 + (a& + 6 2 ), -a 3 & - (- a6 - 6 2 ). 

59. 2.r + (- Sy - 4*), 2x - (3?/ + 4s). 61. 12. 63. 20. 65. 17. 67. 1. 

69. -6. 71. 8. 73. 10. 75. 13. 77. 13. 79. |. 81. —• 

7 14 

EXERCISE 1-4. PAGE 24 

I. 2.n/. 3. x 2 y 2 . 5. 5a 2 - 66 2 . 7. -4a. 9. a: 2 + 2xy. 

II. z 3 + Sx 2 - 3.r + 2. 13. -4xy. 15. 5a: 2 // 2 . 17. 3a 2 + 26 2 . 

19. -2a + 26 - 2c. 21. x 2 - 2xy + y 2 . 23. ^ a: 3 - ^ s 2 + ~ xy + y y*. 

25. -18a;?/. 27. -lOa; 3 ?/ 2 . 29. 24a 4 /> 2 c. 31. -6a; 2 - 4xy. 33. 12x 2 y 2 + Gxy*. 
35. 2a; 2 + xy - 3?y 2 . 37. 4.? 4 - 8x*y - 4.r 2 ?/ 2 . 39. a; 3 - 2.r?/ 2 + 1/ 3 . 41. -2x. 
43. 2x*/. 45. -1. 47. 2. 49. -zy + 2x - 3?/. 51. x + 5. 53. * + y. 
55. s 8 + x 2 y + xy 2 + */ 3 . 57. x 2 - xy + y 2 . 59. .r 2 + .ry + y 2 . 61. 2y - xy. 
63. x 3 + 3x 2 + Sx + 1 ; x 2 + 2x + 1. 65. 2.r 4 - 9a; + 6. 67. Odd values. 

339 



340 Appendix 8 

EXERCISE 1-5. PAGE 30 

1. 3(x + 2y). 3. 2(2x + 7). 5. a(3x + 2). 7. -x(a - 2c + x). 

9. a(x - 2y + 3*). 11. ?/ 2 (5 + 3?/ - a). 13. (x - 4) (x + 4). 

15. (2* - 3) (2x + 3). 17. (7x - 11) (7x -f- 11). 19. (3?/ - a) (3?/ + a). 

21. a 2 x 2 (x - 3a) (x + 3a). 23. (0.1 - 6) (0.1 + b). 

25. (7x*/ - 12a6) (7xy + 12a&). 27. x(6 + x 2 ) (6 - x 2 ). 

29. 2(x + 2) (x 2 - 2x + 4). 31. (xy + z 2 ) (x 2 y 2 - xyz 2 + * 4 ). 

33. (5p 2 a 3 + r 6 ) (25p 4 g 6 - 5p 2 ry 3 r 5 + r 10 ). 

35. (x + y 2 ) (x - ?/ 2 ) (x 4 + x 2 y* + ?/ 8 ). 

37. 3[3x 2 - (2y + 3z)] [9x 4 + 3x 2 (2?/ + 3e) + (2y + 3z) 2 ]. 

39. (x + y - 2i + w) [(x + ?y) 2 + (x -f y) (z - iv) + (z - w) 2 ]. 

41. (6x - y) (36x 2 + Qxy + 2/ 2 ). 43. x(6 - x 2 ) 2 . 45. (xy - 9) 2 . 

47. (x 2 - 5) (x 2 - 3). 49. (4x - 3) (2x + 1). 51. (x - 2y) 2 . 

53. (x - 4) (x + 3). 55. (x 2 + 1) (x 2 + 2). 57. (1 + 15x 3 ) 2 . 

59. 5x(x + 4) (x - 2). 61. (2x + 1) (x - 6). 63. (2x - 1) (x + 3\ 

65. (6x - 1) (x - 6). 67. (x - 0.6) 2 . 69. (a + 3) f.r + 2y). 

71. (4x 2 - 5) (2x - 3). 73. (x + 3) (2a + y - z). 75. (x 2 - 3) (x + 1). 

EXERCISE 1-6. PAGE 33 

1. 2. 3. 1. 5. x - y. 1. 3x 3 ?/. 9. 2xy*z. 11. 1. 13. 24. 15. 24xyz. 
17. 36x«y*z*. 19. x 2 - 4. 21. (x 2 - 49) (x - 3) (x 3 - S) (x 2 - 4). 

EXERCISE 1-7. PAGE 35 

1. (a) 12. (c) 4a 2 . (e) 18x// 3 . (g) x - 1. (i) (a - x) (h + x). 

n , v 101 , v 3x 2 y , v (>x?y 3 , . 1 ,. x 3x - 7 ,, . 3x + 2 

, x x 2 - (y - 3) 2 , x , x 1 , N 3a + 6 , x x 

Wiq^' «»*-Jfo. <«M- < s > 2^zv (u) i+x-2/ 

EXERCISE 1-8. PAGE 38 

- 17 94 „ 57 _ -2 n x 3 - 3x 2 - 6 lt x 2 + 3x - 5 
L 24 # 3 -~2T 5 -~20' 7 -xT3T' 9 ' x 3 = 1 1L ii 

1Q x 2 + x + 4 3x 2 + 12x + 5 1 + x 2 

(x + 1) (x - l) 2 ' (x + 2) (x + 3) (x + 4) ' 1 - x " 

10 - x(x + by) 2(a 2 - ab ± b 2 ) 2x + y±3 9 _ x 2 -xy+ y 2 

19. t ■ r-7 rr • 21. ; j~z • 16* • J5. r r « 

(x + y) (x - i/) 2 a 2 - b 2 xy x 2 - y 2 

EXERCISE 1-9. PAGE 41 

54 x 6 a Q 7a 3 6 4 3x ? /(x - 1) 

(x - 1) (x + 2) (9x 2 + 6x + 4) (2x - 5) (4x + !)(«- 8) 

1A (x - 7) (x - 2) ' 1& - 9x 2 - 4 ' 1/# (5x + 1) (x - 5) ' 



Appendix 8 341 

-3(4* +5) (2s +3), 28. 70. 3^. 

42/M4* 2 + 5) 5 59 5z 

».^ + *. +rt . » ";; T ;_ + T M> - «. l -^£&- 

43. y f + i 2a \ - 45. (*+?)*. 



2/(2/ + &) 



EXERCISE 1-10. PAGE 47 



1. a; = -5/2. 3. x = -3/7. 5. a; = -2. 7. -rr 2 - 9.-11. 11.7. 13.10. 

lo 

15.-2. 17.?/= — ^ 19. 2/= -y-" 21. y=— g— • 

_ 9r 4- 17 —71 —7 

23.?/= , ■ 25.-5. 27.7. 29.1/4. 31.3. 33.-^. 35.-/. 

4 2 4 

9*37 4-^8 

37. =£■• 39. ^. 41. 4. 43. ~- 45. 2. 47. £-. 49. 40, 58. 51. 6, 7. 

46 11 28 41 ' 

53. 921,600 sq ft. 55. 170 adults, 330 children. 57. 20°, 40°, 120°. 

EXERCISE 2-1. PAGE 52 

2. (a) 5. (c) 4. (e) V34. (g) V§9. 

3. (a) 5. (c) V2. (e) 7. (g) 4. (i) v^T&l (k) 2. 

EXERCISE 2-2. PAGE 56 

I. The area of a circle is a function of the radius of the circle, A = 7rr 2 . 

3. The area of a trapezoid is a function of its altitude and bases, A = ~ (&i + W. 

Z 

5. The volume of a cylinder is a function of its height and the radius of its base, 

V = wr 2 h. 
7. The annual premium of a life insurance policy is a function of the applicant's age 

and physical condition, of the type of policy, of the company's rate policy, etc. 

No formula can be written. 

9. -3, -5, 3, -2, 2 V2 -3, -3, 3/4, 2y - 3, \ - 3, ^—g- 

II. - ~ , 288, 6?/ 2 + 7. 13. 0. 15. 14. 17. v 2 + q 2 + r 2 . 19. 0. 21. 30. 
23. 5/7. 25. 0. 27. A = nr', C = 2xr, A = ^- , C = 2 \ArX 

29. £ = ^/36tt F 2 . 31. all .r. 33. all a\ 35. all x. 37. all a;. 39. all x. 
41. all x. 43. all a?. 45. x * 0, 1. 47. | a; | £ 3. 49. a: = 0. 51. all a;. 
53. x * 0, -2. 55. all a\ 57. a: ^ -1. 59. -2^xg2. 

EXERCISE 2-3. PAGE 57 

1. 3,2.3,1*1. 3. 0,0.5. 5. 1,0,1,0,1. 



342 Appendix 6 

EXERCISE 2-4. PAGE 59 

l.lf=yi. 3. 567. 5. 5. 7. 10/3. 9. -2. 13. 9/4, 27/8. 

15. 1000/1, 100/1. 17. 327°C. 19. 99.5 lb, 95.2 lb. 21. 0.0324 in. 

EXERCISE 3-1. PAGE 67 

1. (a) (1, 0). (c) (0, 1). (e) (1, 0). 

2. (a) (0.54, 0.84). (c) (-0.99, 0.14). (e) (-0.65, -0.7G). 

3. (a) V5/2. (c) 1. (e) -1/2. (g) -2. (i) - V3/2. 



7. sin t 


COS t 


tan t cot t sec t 


esc £. 


(a) - 


± V3/2 


± 1/V3 ±VS ± 2/y/S 


2. 


(c) ±5/13 


12/13 


=b 5/12 db 12/5 — 


=fc 13/5. 


(e) ± V3/2 


-1/2 


db V§ ± 1/V3 — 


±2A/3. 


(g) ±l/v<5 


±2/v / 5 


1/2 — ± V5/2 


± Vs. 


(0 - 


±4/5 


± 3/4 ± 4/3 ± 5/4 
EXERCISE 3-3. PAGE 74 


-5/3. 



1. (a) 1. (c) 0.58. (e) 0.86. 

2. (a) 0.9927. (c) 24.52. (e) -1.500. (g) 0.2571. (i) 5.798. (k) 1.011. 

EXERCISE 3-4. PAGE 80 

1. tt/3. 3. tt/6. 5. 2tt/3. 7. tt/15. 9. 4tt/3. 11. 2tt/5. 13. 43tt/36. 
15. 107tt/60. 17. 7tt/20. 19. 0.8090. 21. 1.4358. 23. 3.2107. 25. 1.6323. 
27. 45°. 29. 270°. 31. 15°. 33. 630°. 35. 27°. 37. 470°23'54". 39. 43°43'. 
57. 14tt/3, 1.25 radians. 59. 14.74 in. 
61. (a) 4 radians, (b) 16/9 radians, (c) 0.04 radian. 

EXERCISE 3-5. PAGE 85 

1. 0.5925. 3. 1.092. 5. 0.7412. 7. -1.453. 9. -0.2462. 11. 0.2504. 

13. 1.181. 15. 9.010. 17. 0.6817. 19. 0.8437. 21. 0.9831. 23. 0.5154. 

25.0.2930. 27.-9.462. 29.-1.059. 31. 173°10', 353°10'. 

33. 42°30', 222°30'. 35. 3°10', 183°10'. 37. 35°20 ; , 144°40'. 

39. 56°30', 236°30'. 41. 264°50', 275°10'. 43. 83°10', 263°10'. 

45. 55°20', 235°20'. 47. 42°7', 222°7'. 49. 18°9', 341°51'. 51. 31°35', 328°25 ; . 

53. 67°4', 247°4'. 55. 7°30', 187°30'. 57. 97°36', 262°24'. 59. 93°11', 273°11'. 

61. 0.8016. 63. 3.079. 65. -100.00. 67. 0.3459. 69. -0.7073. 71. 1.214. 

73. 0.220, 6.060. 75. 1.120, 4.260. 77. 0.755, 3.895. 79. 1.158, 5.122. 

81. 1.143, 1.997. 83. 0.574, 5.706. 



EXERCISE 4-1. PAGE 96 

27' - -v. ;. 81. 9. — < 



64 1 

1. 3y. 3. 5=. 5.100. 7.81. 9.—. 11. ay*. 13.1. 



Appendix B 343 

15. x+2(xyy<* +y. 17. ^j-p 19. *"* ~ | - 21. 4(5*' 2 ). 28. (21)»'». 

25# M!!!. 2?# (M^. 29.^. 31.^. 33.(5292)"*. 35. z™ y™. 
V x y 9 v ! u 

Q7 9 +3 \/2 QQ - (1 + yg) • x + V^^9 
37. ? 39. 41. 



43 ( x + y/x 2 - y 2 \ 2 i5 (3 - 2x) \/2x - x 2 < 4? 1 + z 2 - ^/l + x* 



2 -x * s 2 (l + x 2 ) 



49. 2(2 - z 2 ) 3 ' 2 . 



EXERCISE 4-2. PAGE 100 

1. 54. 3. 3/2. 5. 6,652,800. 7. 5/24. 9. n(n - 1). 11. (n + l)n. 

13. ^-^ • 15. — ^r • 19. x 7 - 7x« + 21a; 5 - 35x 4 + 35a; 3 - 21x 2 + 7x - 1. 

21. 8a 6 - 36a 4 6 2 + 54a 2 6 4 - 276 6 . 

23. x* 12 + 10a; 3 ' 2 + 40a; 1 ' 2 + SOar 1 ' 2 + 80ar 3 ' 2 + 32ar 6 ' 2 . 

25.^-4^+6-4^+^. 

27. x 12 - 6a; 10 ?/ 2 + 15z 8 ^ 4 - 20x«y« + 15a; 4 ?/ 8 - 6x 2 y 10 + ?/ 12 . 

29. x 2 + ?/ 2 + z 2 + 2a;?y + 22/z + 2zz. 

31. x 8 + 4a; 7 + 10a; 6 + 16a; 5 + 19a; 4 + 16a; 3 + 10a: 2 + 4x + 1. 5*. -8x 7 . 

35. 7920a 8 6 4 . 37. -14a: 2 . 39. 2 11 • 3 5 • 5 • 7 • 13a; 33 ?/ 8 . 41. 924a: 3 ?/ 3 . 

EXERCISE 5-1. PAGE 104 

1. log 2 8=3. 3. log 3 81 = 4. 5. log, 1000 - 3. 

7. log 256 2 = i • 9. log 100 10 = 0.5. 11. y = log 10 x. 13. 8 2 = 64. 

15. 2-« = ~ • 17. 7 3 = 343. 19. 10 4 = 10,000. 21. 4 3 ' 2 = 8. 23. 6. 25. 2. 

4irr3 93 7 

27. 3. 29. 1/10. 31. 5/2. 33. No solution. 35. log 6 ^- • 37. log5~r- 

39. log* (u - y/u 2 - a 2 ). 

41. (a) log* 7T + 1/2 log 6 Z - 1/2 log* g. (b) 2 log* 7 T + log* g - 2 log* W. 

EXERCISE 5-2. PAGE 107 
1. 1. 3. 4. 5. -1. 7. -3. 9. -1. 11. 5. 13. -1. 15. -1. 17. 7314. 
19. 7.314. 21. 7314000. 23. 0.007314. 

EXERCISE 5-3. PAGE 110 

1. 1.5441. 3. 2.0212. 5. 7.7931-10. 7. 4.3636. 9. 9.5490-10. 11. 8.8215-10. 

13. 9.9279-10. 15. 0.4536. 17. 7.8452-10. 19. 9.8908-10. 21. 0.4972. 

23. 8.9439-10. 25. 9.9824-10. 27. 9.9476-10. 29. 9.1306-10. 31. 46.4. 

33. 0.262. 35. 504. 37. 0.0000000000276. 39. 69.2. 41. 292.3. 

43. 5,454,000,000. 45. 0.06114. 47. 4.554. 49. 0.00001072. 51. 0.6021. 

53. 0.4266. 55. 1.585. 57. 3.728. 59. 2.5023. 61. 1.6297. 



344 Appendix B 

EXERCISE 5-4. PAGE 112 

1. 8.540. 3. 0.04292. 5. 3.183. 7. 0.0008416. 9. 0.1104. 11. 54.61. 
13. 48.91. 15. 11,670. 17. 0.1795. 19. 0.02950. 21. 20.56. 23. 538,100. 
25.1.708. 27.-1.021. 29.1.249. 31.0.4343,0.2171,9.5657-10,23.1,22.46. 
33. 127,900,000 sq ft. 35. 12.62 ft. 37. 6,070,000 sq ft. 39. 16.5 amp. 
41. $1,074.00. 

EXERCISE 5-5. PAGE 114 

1. 2.3026. 3. 1.4429. 5. 0.8735. 7. 6.0001. 9. 1.5373. 11. 2.0794. 
13. 1.4307. 15. 0.8228. 

EXERCISE 6-1. PAGE 119 

1. B = 57°, b = 18, c = 22. 3. A = 18°50', a = 21, c = 66. 
5. A = 27°!', B = 62°59', c = 7.012. 7. A = 22°44', b = 10.30, c = 11.17. 
9. B = 53°39', b = 13.40, c = 16.64. 11. £ = 46°43', a = 73.66, c = 107.5. 
13. A = 8°58', 5 = 81°2', c = 793.0. 15. A = 49°3', a = 2.663, c = 3.528. 
17. h = 29 ft, I = 33 ft. 19. 113 ft. 21. 227 ft. 23. 13°34'. 25. 4 ft. 

EXERCISE 6-2. PAGE 124 

1. 63°26', 11,000 ft. 3. 60°. 5. N11°24'W, 74 nautical miles. 7. 400 ft. 
9. 80 ft, 173 ft. 

EXERCISE 6-3. PAGE 131 

1. (a) 5. (c) 2 VS. (e) yft. (g) 4. 

2. (a) 5 [3/5, 4/5]. <c) 2 V5 [- ^, ^] • (e)V^,^]. 
(g) 4 [0, 1]. 

3. (a) 5, 53°8'. (c) 5, 143°8'. 

4. (a) VS, 63°26'. (c) V73, 159°27'. 

5. 12, 0°. 7. 22 knots, 20 knots. 9. 60 lb. 11. 49 lb, 250°45 / . 

EXERCISE 6-4. PAGE 133 

I. 9.8733-10. 3. 8.8059-10. 5. 0.7391. 7. 9.3661-10. 9. 9.9427-10. 

II. 9.5906-10. 13. 0.0030. 15. 0.1004. 17. 1°20', 181°20'. 19. 8°30', 171°30'. 
21. 18°, 198°. 23. 54°16', 234°16'. 25. 64°2', 295°58'. 27. 30°11', 210°11'. 
29. 53°40', 233°40'. 31. 27°24', 152°36'. 33. 1°27', 181°27'. 35. 65°47', 294°13'. 

EXERCISE 6-5. PAGE 134 

1. B = 59°, 6 = 60, c = 117. 3. B = 64°40', b = 133.9, c = 148.2. 
5. A = 71°15', b = 2.01, a = 5.94. 7. A = 62°51', b = 18.53, c = 40.61. 
9. 0°20'. 11. 671 lb, 200°40'. 13. 607 mph, N36°42'W. 
15. 1815 lb, 1962 lb. 17. 49.19 in. 19. 16,900 lb. 



Appendix B 345 

EXERCISE 7-1. PAGE 138 

l.\(V6-V2). 3.f- 5. 2 -VS. 7. 2 -VS. •.-%$-£' 

11. -3. 13. J (V5 -2 V2), J (1 + 2 a/6). 15. 4/5, -3/5. 

D O 

17. 33/56, 63/16. 

EXERCISE 7-2. PAGE 142 

I. i j/Tv5. 3. ^. 5. | ^2 + v^. 7. 1 + V& 

n / x 120 /ux V26 , . 120 , AS 119 , x 5\/26 , t . 119 
»•(•)- 189. W^e", W-Tl9> W"W (e) ^6~' (f) l69' 

r \ ?o^ m^ 2856Q 

W 2197' W 239 

II. (a) -3/5, (b) -^ |/50+5Vl0, (c) 3/4, (d) 4/3, 

(e)^|/50-5 v / 10, (f)-4/5, (g) ^~, 00 y- 
13. cos 30. 15. tan 60. 17. sin 2 20. 19. cos 20. 21. tan 20. 23. cos 0. 

25. cot £ • 27. tan* ~ • 
4 2 

EXERCISE 7-3. PAGE 144 

1. i [sin 70 - sin 0]. 3. sin 100 + sin 20. 5. | [cos 60 + cos 20]. 

7. _ I [cos 48° - cos 8°]. 9. - £ [cos 60 - cos 40]. 11. 2 sin ^ cos |. 

(\Q Of} 

13. 2 sin y cos y • 15. - 2 sin 50° sin 30°. 17. 2 sin 32°30' cos 7°30'. 

19. 2 sin 43° cos 3°. 

EXERCISE 8-1. PAGE 154 

1. 2tt, 3, 0°. 3. 2tt, 1/2, 0°. 5. 8tt/3, 1/3, 0°. 7. 5tt/2, «, 0°. 

9. 8 radians, 1, 0°. 11. 2?r/5, 3, 0°. 13. tt/6, «, 0°. 15. ir/4, «, 0°. 

4 7 

17. 1 radian, », — radians. 19. 2 radians, <» ; — radians. 21. 2x, 1, 0°. 

23. 7r, 3, ^9 radian. 

EXERCISE 8-2. PAGE 161 

I. s=?fiL±*. 3. s=^-=~. 5. x = -12t/ -22. 7. tt/6. 9. *r/6. 

II. ir/2. 13. tt/3. 15. tt/6. 17. -tt/3. 19. y . 21. -24°27'. 

23. 12/5. 25. 5/13. 27. 3/4. 29. 0.3919. 31. ^ • 33. 3/4. 35. -3/5. 
37. -tt/2. 39. -w. 41. w. 43. u. 45. Vl±Jf!. 47. Ut 49. w . 51. i/ u . 



53. ^\ U \ - 55. \/T^uK 
1 - ti 2 



346 Appendix 6 

EXERCISE 9-1. PAGE 168 

1. x = 10/7, y = -6/7. 3. rr = -9/7, 2/ = 15/7. 8. a? = -2, */ = 1. 

f 18 - 1 o ~ 47 1 11 - 10 <w* 

7 -* = 18 ,y= uT' 8 -* = "20"^ = 20 - ".a; =—,*/= 9/7. 

** -I 1 68 ir T • j. i. iit - 21 T/rr ~22 

13, a; = ~7F"; 2/ =: tH • 15. Inconsistent. 17. # = — — , y = 7/5, 2; = — £— • 

19 - * =H * = -^ * =§' 21 * * = 10/7, y = 25/7, * = ~ 1/7 * 

23. x = 16/11, y = -21/11, 2 = -9/2. 25. a = 9/5, y = -1/5, z = 1. 

27. x = 10/7, y = 4, 2; = 20/3. 29. 3 = 0, 2/ = 5/7. 31. Inconsistent. 

33. Inconsistent. 35. Consistent and dependent. 

37. Consistent and independent. 39. Inconsistent. 

41. Inconsistent unless c = 1. Consistent and dependent if c = 1. 

EXERCISE 9-2. PAGE 172 

1. (1/4, 0), (0, -1/3). 3. (4, 0), (0, -4). 5. (0, 0). 7. (-4/3, 0), (0, 4). 
9. (5/3, 0), (0, -5). 11. x = 6/5, y = 4/5. 13. x = 23/7, ?/ = 22/7. 

15. x = 1, y = 5/8. 17. * = -j|,y=?- 

EXERCISE 10-1. PAGE 180 

1. 5. 3. 0. 5. 0. 7. 7. 9. -11. 11. 22. 13. x =4.1, 1/ =0.3. 
15. x = 25/13, ?/ = -5/13. 17. (y, 0), (0, -17). 
19. (-3, -1), (2,4), (6, -3). 

EXERCISE 10-2. PAGE 185 

1. x = 2, y = 3, z = -2. 3. x = 1, y = -2, z = 2/3. 

5. a: = 2, */ = 3, z = —2. 7. x = — 8z, 1/ = -3z. 9. No nontrivial solution. 

EXERCISE 10-3. PAGE 187 
1. 0. 3. 0. 5. -110. 7. 308. 9. 2184. 

EXERCISE 11-1. PAGE 193 

I. + 4t, - 4i. 3. - 3r, + Si. 5. - 6\a\i, + 6|«|i. 
7. 3 \/2 + 3 V§», 3 \/2 -3 V^'. 9. 1 + 4 \/2i, I - 4 V§*. 

II. y/\b ^ S\a\y/^ i, V^ -^\a\y/ob i. 13. -t. 15. -i. 17. -1. 
19. f. 21. 1. 23. 0. 25. 0. 27. * = 3/2, y = 1/3. 29. x = 6, y = -5. 
31. 3 = 1, y = -4. 33. x = 2, y = 5/2. 35. x = 7/3, y = 4/3. 37. 7 + 3t. 

39. I + ^ i. 41. -1 + K. 43. 25 + Oi. 45. -1 + i. 47. 1 + Ot. 



Appendix B 347 

49. -3 + (\/2 + y/2)i. 51. 6 + Oi. 53. 13 + lit. 55. -8 + 4i. 

57. 11 - 3*. 59. 27 + 24*. 61. 28 + 16i. 63. 5 - 2 V§ + (2 + 5 y/2)i. 

65. -2 - 26*. 67. ~ + I *. 69. ~ - -£■ *. 71. 33 - 22i. 73. ~ + £ i. 

J Z 41 41 Dl Dl 

7 - 3 4. 77 025 1019 . 
75- " 25 + 25 *' 77- 3233 + 3233 *" 

EXERCISE 11-2. PAGE 197 

13. 3 - 2*. 15. -1 - 9*. 17. »2 + 3*. 19. 6 - 4*. 21. + 3*. 

23. 5 + (2 - y/Z)i. 25. V2(cos 15° + i sin 45°). 27. 3(cos 270° + * sin 270°). 

29. ~-(cos 54°44' + * sin 54°44'). 31. 13(cos 67 l 23' + i sin 67°23'). 

33. 3 (cos 0° + * sin 0°). 35. ^p (cos 320°12' + * sin 320°12'). 
EXERCISE 11-3. PAGE 202 

1. (i + VS) + (1 - VS)«. 8.-2V3+K. 5. (1 "" 2 V§) + (1+ 2 ^ 5) <. 
7.1+1. 9.-1. 11. -1/2-^*. 13.^+^*. 16. 1 +|t. 
17.-2"*. 19. — (0.6065 + 0.7951 *"). 21.-1. 

23. 2[cos (9° + A' 90°) + * sin (9° + A; 90°)], ft = 0, 1, 2, 3. 

25. 2[cos (36° + ft 72°) + i (30° + ft 72°)], ft = 0, 1, 2, 3, 4. 27. 1, i, -1, -t. 

29. cos(S0° + ft 120°) + i sin (80° + ft 120°), ft = 0, 1, 2. 

Q1 14 + 18* QQ 828 + 154* , 

31. amperes. 33. ~=^ ohms. 

o 17o 

EXERCISE 12-1. PAGE 206 

1. 0, -7. 3. -1/2, -3/4. 5. 3, -2. 7. 3,3, -3. 
9. sin = 0, 1; 0°, 90°, 180°. 11. sin = 1, -2; 90°. 
13. sec 0=2, -8; 00°, 300°, 97°11', 262°49'. 
15. cot = -3, 2; 161°34', 341°34', 2(>°34', 206°34'. 
17. cot 0=7, -17, 8°8', 188°S', 17G°38', 356°38'. 
19. sin = 0, 2, - | ; 0°, 180°. 

EXERCISE 12-2. PAGE 209 

I. 10, -2. 3. 10, -3. 5. - 1= ^ . 7. ^if^i. 
9. tan = 1 ± V5; « = 67°30 ; , 247°30', 157°30', 337°30 / . 

II. sec = 1, -3; 0°, 109°28', 250°32 / . 



348 Appendix B 

13. esc 6 = * ± 4 V ^ ; 36 Q 23', 143°37', 237°30', 302°30'. 

15. cos $ = 0.4142, -2.414, 65°32', 294°28'. 17. (a - 3) 2 + 4(y + 2) 2 = 4. 

19. (s - 5) 2 + My - 5)2 = 16. 21. 4(s - 2) 2 + 9(i/ - l) 2 = 36. 

23. 4(rc + 4) 2 - 9(y - 2) 2 = -36. 25. 3(x + 10/3) 2 - (y + l) 2 = 64/3. 



27. \/2[(s -4) 2 +9/2]. 29. . 31. [2((rc + 7) 2 - 32)]~" 2 . 

y/{z — 3) 2 - 16 
33. [9((a; + 4/3) a + l)]-i/3. 

EXERCISE 12-3. PAGE 212 

I. 1, -7. 3. -1 ± >/6. 5. -1, -1. 7. - 3 ^ v7i < 9 5/2 ^ _ 1/2> 

II. -1, - 11/8. 13. -1, -15/7. 

15. tan = 1, -5/2, 45°, 225°, 111°48', 291°48'. 
17. sin 6 = 3.7913, -0.7913; 232°18', 307°42'. 

19. cos = r ; no values of 0. 

o 

21. sec = 0.414, -2.414; 114°28', 245°32'. 

23. esc 6 = -0.6972, -4.3028; 193°26', 346°34'. 

EXERCISE 12-4. PAGE 213 

1. 18. 3. -4. 5. 8 ± 4 \/5. 7. 5. 9. *izi|-^Jl . n # l, 5. 

EXERCISE 12-5. PAGE 215 

1. d= a/5, ±2i. 3. ±», ±t VO. 5. ±2, ±3. 7. 1 ^ 2l • 
9. -4 ± VT5, 3±2y^. 

EXERCISE 12-6. PAGE 216 

1. Conjugate imaginary. 3. Heal, unequal, irrational. 
5. Real, unequal, irrational. 7. Real, unequal, rational. 
9. Real, equal, rational. 11. Conjugate imaginary. 
13. Real, unequal, irrational. 15. Conjugate imaginary. 

EXERCISE 12-7. PAGE 218 



1.-2,-1. 3.0,2. 5.3/2,6/5. 7.6/5,-1/5. 

9. |> ~~p 11. a* - 2x = 0. 13. a 2 - 9x + 18 = 

17.3 2 + 1=0. 19. z 2 - (\^ - Vl)x = 0. 21. z 2 ~2V§z+8 = 0. 



'•f'lffiT 11. » 2 -2x=0. 13. a; 2 - 9x + 18 = 0. 15. s 2 - 4a; + 9 = 0. 



Appendix B 349 

EXERCISE 12-8. PAGE 224 

1. Circle. 3. Intersecting lines. 5. Hyperbola. 7. Parabola. 
9. Intersecting lines. 11. Hyperbola. 13. Intersecting lines. 
15. Intersecting lines. 17. Intersecting lines. 19. Parallel lines. 

EXERCISE 12-9. PAGE 227 

1. (3, 3), (-3/2, 3/4). 3. (3, 4), (-4, -3). 5. (2, 4), (-3, 9). 
7. (3, 2), (-1, -6). 9. (2, 3). 11. (± 2 V§, 1). 
13. No solution. 15. (±4, ±2). 

EXERCISE 12-10. PAGE 231 

1. (3, 3), (-3/2, 3/4). 3. (3, 4), (-4, -3). 5. (4, -3), (-2, 6). 

7. (4, 0), (-5, 3). 9. (4, 6), (-3, -1). 11. (±3, I) (±3 yAdi, -10). 

13. (*^;,*^). 15. (±2, ±4). 17. (1,2), (2,1). 

19. (4,1), (-4,-1), (14, -4), (-14,4). 

21. (4, 2), (-4, -2), (V6, -2 V6,), (- V6, 2 ^6). 23. (5, 5), (-5, -5). 

25. (3, 4) (4, 3), (-3, -4), (-4, -3). 

27. (2 + i V2, 2 - i y/2), ( -3 + i \fl, -3 - i y/l), (2 - i \/2 y 2 + % y/2), 
(-3 -iV7, -3 +iV7). 
/ 31 -3 V93 31+3 \/93 \ / 31 + 3 \/93 31-3 \/93 \ 

^- v 31 ' 31 ; ' v 31 ' 3i~; * 

EXERCISE 12-11. PAGE 233 

1. 6. 3. 4/3. 5. 2.292. 7. 1.682. 9. 1. 11. 1.836. 13. 2.718. 15. 8.547. 

17. 2.944. 19. 49.3. 21. 10, 0.1. 23. 0.7S74. 25. -0.44. 
27 logO -ac) -log 6- log c , 2d ^ {y±v - r — ih , La8M4> 
log c 

EXERCISE 13-1. PAGE 239 

I. Q : x - 7, # : 0. 3. Q : x - 1, 72 : 2. 5. Q : x 3 - 3.r 2 + 6x - 24, # : 78. 
7. Q : 2z 3 + 3x 2 + 4, 72 : 0. 9. Q : x 2 + 5x + 8, R : 11. 

II. Q : x 2 + 2x - 15, R : 0. 13. Q : a; 3 + 3a; - 6, R : 0. 

15. Q : x**- 1 + x*~ 2 y + • • • + y n ~ l , R : 0. 17. 52, 2. 19. 24, -36. 

EXERCISE 13-2. PAGE 241 

1. 2a; 2 - 2x - 3. 3. Not a factor. 5. x 2 -f 2x - 5. 
7. x 7 + 2xHj 2 + 4.rV -J- 8*V + 16x*y* + 32x 2 y 10 + 64.ri/ ,a -f 128y". 
9. Not a factor. 11. hx 2 + 2ab(5 - Sab 2 )x - 12a 3 6 4 . 13. Not a factor. 
15. 12x 3 - 22a: 2 - 34a; + 60. 17. 24a; 3 - 90a; 2 + 39a; + 45. 



350 Appendix B 

EXERCISE 13-3. PAGE 244 

1. 1 ± V&, -3. 3. 1 =fc i, ±3. 5. 2 ± V2i, 1, 1, 2. 
7. s 3 - ix 2 + 9x - 10 = 0. 

EXERCISE 13-4. PAGE 247 

1. -1/3, 1/2, 5/3. 9. -1. 11. 2, 2, -2, -2. 13. 2/3, ~* y^ - 
15. 1/2, =bi. 17. 3/5, 1 =fc i. 19. 1/2, ± V5- 

EXERCISE 14-1. PAGE 253 

I. x < 3. 3. a: > -5. 5. x < 4. 7. a; < -1/2. 9. x < -1. 

II. -\<*<\- M.-§<*<|« 15 --§ <a;< jr 17. ~13<x<13. 
19. -4 £ x £ 4. 21. -1 < x < 1/3. 23. x < -5/3, x > 2. 

25. No values of s. 27. x < -3, -2 < x < -1. 29. 1 < a < 2, a; > 3. 
31. | x | ^ 5. 33. x < -2, s > 0. 35. 3 < -1/2. 

EXERCISE 15-1. PAGE 260 

1.1,2,4,7,11. 3.1,2,3,5,11,35. 7. 2, 5, 11, 23, 47; 2, 7, 18, 41, 88. 

EXERCISE 15-2. PAGE 263 

1. 17, 20. 3. 18, 25. 5. Not an arithmetic progression. 7. 46 — 3a, 5b — 4a. 

9. 4fl = 26 , 5a ~ 36 . 11. f 26 = 78, £ 26 = 1053. 13. Iio = 10, S i0 = 55. 

15. hi = 149, <S 75 = 5625. 17. J 20 = 5.9, S 20 = 61. 19. i,oo = 100, S, oo =505O. 

21. l l0 = 100, S,o = 550. 23. a = - y , i 4 « = y • 

25. a = 1, n = 13 or a = -1/2, n = 16. 27. /« = 9, n = 9. 29. n 2 . 

31. d = -^°- • 33. 33, 16. 35. d = - ~* • 37. $37.75. 39. 282. 
11 ' a(k + 1) 

EXERCISE 15-3. PAGE 264 

1. 1/15, 1/19. 3. 1/20, 1/25. 5. 4, 16/5. 7. 1/47. 9. j, | , y , |. 
„ 120 60 120 , 

TT'T'TT' 

EXERCISE 15-4. PAGE 267 

I. 128, 512. 3. 256, 1024. 5. 8, 16/3. 7. ^, ^. 9. -1, 1. 

II. l„ = P , 5„ = 9[1 - (2/3)"]. 13. J 101 = 10-", 5 101 = yp [1 + 10—>]. 
15./.=^,^=^^. 17.3,-14. 19.^51. 21. ±6, 15/2. 



Appendix B 351 

23. 10; 100; 1000; 10,000; 100,000. 

1 *■-• x n fix n 
27. As n increases, the sum approaches 3 as a limit. 29. 



(1 - a;) 2 1 - x 
EXERCISE 15-5. PAGE 270 

1.64/65. 3. 9 + 2 5# 16 ' 7# 3 - 9 - 5/8 - 1L 2/U - 

13. 36, 36 [1 - (1/3) 20 ], 36(1/3)20. 15. 12 ft, 12 [1 - (3/4) 10 ] feet. 17. 1/9. 

19.10/11. 21.1/6. 23.3. 25. g^- 

EXERCISE 15-6. PAGE VT2 

1.1-*+-—. 3 -i+l"f+S- 8 - 1 -5 + X-S- 
7 -i-S+S"S- «-i-S + SP-^- »• 1-2190. 13.1.3684. 

X X 2 X 3 X* X 3 X* X* x° 

15. 0.8508. 17. 0.9415. 19. 1.0149. 

EXERCISE 17-1. PAGE 279 

1. 72. 3. 216. 5. 9999. 7. 20,160; 7560. 

EXERCISE 17-2. PAGE 282 

1. 20; 210; 95,040; 143,640; 970,200. 3. (a) 720, (b) 48, (c) 480. 5. 125. 

7. 8 Co<z 8 4- sCitfb + 8 C 2 a 6 6 2 + 8 C 3 a*b* + 8 C 4 a<6 4 + 8 C 5 a 3 6 6 + 8 C 6 a 2 6« + 

sC 7 ab 7 -f 8 C 8 6 8 . 
9. 9,979,200. 11. 66. 13. 63. 15. 52 " 



(13!) 4 



EXERCISE 17-3. PAGE 288 

1. 2/5, $24.00. 3. 1/3, 2/3. 5. $7.29. 7. $2.42. 9. 0.553. 11. 16/63, 

iQ JL 1* A 215 

16. 1728 - 15. 72 , 216 ' 

EXERCISE 18-1. PAGE 295 

1. C = 9S°5S', 6 = 14.55, c = 20.46. 3. A = SOW, a = 1169, 6 = 1079. 
5. B = 5°31', 6 = 1.051, c = 7.513. 7. A = 17°3', £ = 100°26', b = 71.18. 
9. B = 24°27', C = 101 °20', c = 1193. 11. No solution. 13. 615.3 ft. 
15. 449 ft. 17. 12.6 in. 19. 3158 lb, 101°37'. 

EXERCISE 18-2. PAGE 298 

I. c = 270, B = 51°30', A = 69°50'. 3. a = 290, B = 11°50', C = 101°. 
5. a = 100, £ = 10°, C = 20°. 7. 48°20'. 9. No solution. 

II. A = 54°20', £ = SOW, C = 66°. 13. 3257 ft. 15. 700 ft. 

17. c = 20, A = 63°, B = 73°, C = 44°. 



352 Appendix 8 

EXERCISE 18-3. PAGE 301 

1. A = 50°, B = 70°, c = 56. 3. A = 61°, C = 23°50', 6 = 262. 
5. B = 31°11', C = 70°, a = 79.25. 7. A = 47°, £ = 58°, C = 75°. 
9. A = 42°20', B = 57°30', C = 80°10'. 11. 4 in., 5 in. 13. 11 in. 
15. 55°30', 59°50', 64°40'. 

EXERCISE 18-4. PAGE 304 

1. 88.41. 3. 243,900. 5. 9285. 7. 90. 11. 54 ft. 15. 45 ft. 
17. 38 sq ft, 12 sq ft. 



Index 



Abscissa, 50 
Absolute value, 14, 57 
Absolute value function, 57 
Addition, 2 

of algebraic expressions, 21 

of fractions, 36 

fundamental laws for, 2 

of real numbers, 2 
Addition formulas, 135 
Algebraic expressions, 17 

addition and subtraction of, 21 

division of, 23 

multiplication of, 22 

symbols of grouping of, 18 

transposing terms of, 43 

Amplitude of a trigonometric 

function, 149, 152 
Angle 

coterminal, 76 

definition of, 75 

degree measure of, 77 

of depression, 120 

of elevation, 120 

initial side of, 75 

measurement of, 76 

negative of, 75 

positive of, 75 

radian measure of, 77 

standard position of, 75 

terminal side of, 75 

trigonometric functions of, 81, 82 

vertex of, 75 
Antilogarithms, 109 
Arc of a circle, 78 
Area 

of a sector of a circle, 78 

of a triangle, 302 
Arithmetic means, 262 
Associative law 

for addition, 2 

for multiplication, 2 
Axes, coordinate, 49 
Axis of symmetry of a parabola, 219 

Base 

change of logarithmic, 113 

exponent and, 16, 86 

of logarithms, 101, 113 
Bearing in navigation and 

surveying, 121 
Binomial, 17 
Binomial series, 271 



Binomial theorem, 97 
combinations applied to, 282 
general term of, 99 
proof of, for positive integral 
exponents, 275 

Braces, 18 

Brackets, 18 

Cartesian coordinates, 50 
Characteristic of a logarithm, 106 
Characteristic of the general 

quadratic equation, 222 
Circular system, 77 
Classification of functions, 61 
Coefficient, 17, 18 
Combinations, 278, 281 

and the binomial coefficients, 282 
Common difference, 260 
Common logarithms, 105 
Common ratio, 265 
Commutative law, 2 
Completing the square, 206 
Complex fractions, 40 
Complex numbers, 189 

absolute value of, 196 

addition and subtraction of, 192 

amplitude of, 196 

argument of, 196 

congugate of, 191 

definition of, 189 

De Moivre's Theorem, 199 

division of, 193, 198 

graphical representation of, 195 

modulus of, 196 

multiplication of, 192, 198 

roots of, 200 

trigonometric representation 
of, 196 
Composite number, 25 
Composite (reducible) polynomial, 26 
Conditional equation, 18 
Constant, 53 
Constant function, 53 
Convergence of series, 259 

Coordinate systems 

Cartesian, 50 

one-dimensional, 49 

rectangular, 49 
Coordinates, 49, 50 
Cosecant, definition of, 65 



353 



354 



Index 



Cosine, definition of, 64 
Cotangent, definition of, 65 
Coterminal angles, 76 
Counting numbers, 1 

Defective equations, 45 
Degree 

of a polynomial, 17 

relation to radian, 77 

of a term, 17 

term of highest, 17 

unit of angle measure, 77 
De Moivre's Theorem, 199 
Dependent events, 286 
Descartes, Rene, 50 
Determinants, 173 

expansion of, 176 

minors and cof actors of, 175 

principal and secondary diagonal 
of, 174 

properties of, 177 

of the second order, 173 

solution of linear equations by 
means of, 181, 183, 184 

sum and product of, 185 

of the third order, 175 
Directed line segment, 122 
Discriminant, 215 
Distance between points, 50, 51 
Distributive law, 3 
Division, 5, 16, 23, 25, 39 
Domain of a function, 53 
Double-angle formulas, 139 

Empirical probability, 284 
Equality symbol, 12 
Equations 

conditional, 18 

consistent, 163, 164, 165 

defective, 45 

definition of, 18 

dependent, 165 

equivalent, 43 

extraneous roots of, 45, 213 

inconsistent, 163, 165 

independent, 164 

linear, 42, 43, 163, 166, 167 

in quadratic form, 204, 214 

solution of, 42, 163, 171, 181, 183, 
184, 212, 224, 227 
Exponential equations, 231 
Exponents, 16, 86, 88, 90, 92 

Factor theorem, 240 
Factorial symbol, 97 
Factoring, 25, 27 
Fractions, 9 
addition and substraction of, 36 



Fractions (Conh) 
complex, 40 

fundamental operations on, 9, 10 
multiplication and division of, 39 
reduction of, 33 
signs associated with, 34 

Functions 

absolute value, 57 

algebraic, 61 

classification of, 61 

constant, 53 

definition of, 53 

domain of, 53 

exponential, 233 

graphs of, 146, 147, 148, 169, 218, 
233, 244 

greatest integer, 57 

identity, 53 

inverse, 155 

irrational, 62 

linear, 54, 169 

logarithmic, 233 

multiple-valued, 53 

notation for, 55 

periodic, 149 

point, 63 

polynomial, 61 

range of, 53 

rational, 61 

rule of correspondence of, 53 

single-valued, 53 

transcendental, 61 

trigonometric, 63 
Fundamental assumptions, 1 
Fundamental operations, 1, 9, 10, 20 
Fundamental theorem of algebra, 241 

General term in the binomial 

expansion, 99 
Graphs 
of exponential functions, 233 
of inverse trigonometric functions, 

158, 159 
of linear functions, 169 
of logarithmic functions, 233 
of polynomials for large values 

of x, 244 
of trigonometric functions, 146, 

147, 148, 149, 151, 152 

Greater-than symbol, 12 
Greatest common divisor, 30 
Greatest-integer function, 57 

Half -angle formulas, 140, 300 
Highest common factor, 31 
Highest degree term, 17 
Horizontal line, 50, 169 

Identity, 18 
Independent events, 286 



Index 



355 



Inequalities, 248 

absolute, 248, 254 

conditional, 248, 249 

involving absolute values, 14 

properties of, 248 

solution of conditional, 249 
Inequality symbol, 12 
Initial side, 75 
Intercepts, 170 
Interpolation, 83, 108 
Inverse functions, 155 
Irrational functions, 62 
Irrational number, 1 
Irreducible polynomial, 26 

Law of cosines, 296 
Law of sines, 289 
Law of tangents, 298 
Law(s) of exponents, 86 
Least common multiple, 32 
Less-than symbol, 12 . 
Like terms, 21 
Limit of sequence, 258, 259 

Line 

horizontal, 50, 169 

vertical, 51, 169 
Linear equation, 42 

graphs of, 163, 169, 171 

in one unknown, 43 
Literal parts, 17 
Logarithmic computation, 110 
Logarithmic equations, 231 
Logarithms, 101 

base of, 101 

change of base, 113 

characteristic of, 106 

common, 105 

computation by, 110 

definition of, 101 

laws of, 102 

mantissa of, 106, 108 

Napierian (natural), 105 

tables of, 108 

of trigonometric functions, 131 

Mantissa of a logarithm, 106 
Mathematical expectation, 284 
Mathematical induction, 273 
Mean 

arithmetic, 262 

geometric, 267 

harmonic, 264 

Meaning of a m/n , 94 
Meaning of a , 98 
Measurement of angles, 76 
Monic polynomial, 18 



Most probable number, 284 
Multinomial, 17 
Multiplication 

of algebraic expressions, 22 

of fractions, 39 

fundamental laws for, 2 
Mutually exclusive events, 285 

Natural (Napierian) logarithms, 105 
Negative exponents, 90 
Negative numbers, 7, 8 
Number 

algebraic, 62 

complex, 189 

irrational, 1 

negative, 1, 7, 8 

positive, 1, 7, 8 

prime, 25 

real, 1 

transcendental, 62 
Number scale, 6, 49 

One-to-one correspondence, 49 
Operations 

on fractions, 9, 10 

with zero, 5 
Order of fundamental operations, 20 
Order relations for real numbers, 12 
Ordered number pairs, 50 
Ordinate, 50 
Origin, 6, 49 

Parentheses, 18 

Period, 150 

Periodicity, 149, 150 

Permutations, 278, 279 

Phase, 149, 152 

Point function P (t), 63 

Polynomial, 17, 18, 26, 61 

Positive integer, 1 

Positive integral exponents, 16, 8G 

Positive numbers, 7, 8 

Power, 16, 17, 86, 87, 88 

Prime, 25, 26 

Principal branch, 158 

Principal value, 158, 159, 160 

Probability, 278 

empirical, 284 

mathematical, 283 
Product, definition of, 2 
Product formulas, 143 
Progressions, 256 

arithmetic, 260 

geometric, 265 
armonic, 264 
infinite geometric, 268 



356 



Index 



Projections, 122 

Quadrant, 49 
Quadratic equations 

methods for solving, 204, 209 

in one unknown, 204 

in two unknowns, 221, 224, 227 
Quotient, 5 

Radian, 77 

Radius vector, 52 

Range of a function, 53 

Rational exponents, 92 

Rational function, 61 

Rational number, 1 

Real number, 1 

Reciprocal, 4, 5 

Rectangular coordinates, 49, 50 

Redundant equation, 45 

Remainder theorem, 240 

Repeated trials, 287 

Repeating decimals, 269 

Root of an equation, 42, 45, 213, 217 

Scalar quantities, 125 

Scientific notation, 92 

Secant, definition of, 65 

Second degree equation, 204 

Segment, line, 122 

Sequences, 256, 258, 259 

Series, 256, 257, 259 

Sexagesimal system, 77 

Simultaneous equations, 163, 171, 
181, 224, 227 

Sine, definition of, 64 

Special products, 22 

Standard position of an angle, 75 

Sum, definition of , 2 

Symbols of grouping, 18 

Synthetic division, 235 



Tables 

of logarithms, 108 

of trigonometric functions, 71, §2 
Tangent, definition of, 64 
Terminal side, 75 

Terms of algebraic expressions, 2, 17 
Theory of equations, 235 
Transcendental functions, 62 
Triangles, 115 

solution of general, 289 

solution of right, 117, 133 
Trigonometric functions, 63 

of angles, 81, 116 

definitions of, 64, 65 

graphs of, 146, 147, 148, 149, 
151, 152 

of important special numbers, 65 

inverse, 146, 156, 158, 159, 160 

logarithms of, 131 

of sums and differences, 135 

variation of, 146 

Trigonometric identities, 68 

Variable, 53 

dependent, 53 

independent, 53 
Variation, 57, 58 
Vector, 125 

components of, 125 

magnitude of, 126 

multiplication of, by a scalar, 126 

normalization of, 127 

projection of, 125 

representation of, 125, 128 

sums and differences of, 127, 129 
Vertex of an angle, 75 
Vertical lines, 51, 169 

a?-axis, 49 
x-coordinate, 50 

y-axis, 49 
^/-coordinate, 50 

Zero, 3, 5 

Zero polynomial, 61